Dual Nature of Radiation and Matter Test - 10

Given:

Energy of photons \(=\mathrm{h\nu}\)

Work function of surface \(=\mathrm{E_0}\)

Maximum kinetic energy \(=\mathrm{K}\)

Now, frequency is doubled

The energy of photon is used in liberating the electron from metal surface and in imparting the kinetic energy to emitted photoelectron.

According to Einstein's photoelectric effect energy of photon = KE photoelectron + Work function of metal i.e.,

\(\mathrm{h\nu=\frac{1}{2} m v^{2}+E_{0}}\) 

\(\mathrm{h \nu=K_{\max }+E_{0}}\)

\(\mathrm{ K_{\max }= h \nu-E_{0}} \ldots \ldots\) (i)

Now, we have given,

\(\nu'=2 \nu\)

Therefore, \(\mathrm{K_{\max }=h(2\nu)-E_{0}}\)

\(\mathrm{K_{\max }^{\prime}=2 h \nu-E_{0}}\)

From Equation (i) and (ii), we have

\(\mathrm{K'_{\max }=2\left(K_{\max }+E_{0}\right)-E_{0}}\)

\(=2 \mathrm{K_{\max }+E_{0}}\)

\(=\mathrm{K_{\max }+\left(K_{\max }+E_{0}\right)}\)

\(=\mathrm{K_{\max }+h \nu}\quad\) [From Equation (i)]

\(\mathrm{K_{\max }=K}\)

\(\mathrm{\therefore K'_{\max }=K+h\nu}\)

The photoelectric emission is an instantaneous process. The time lag between the incidence of radiations and emission of photoelectrons is very small, less than even \(10^{-8}\) second.

Hence, the correct option is (D).

Given that the threshold frequency for metal surface corresponds to an energy \((\nu)=6.2 {eV}\)

Stopping potential for the radiation incident on it is, \(V_{0}=5 {~V}\)

We know that in the photoelectric experiment

\(E=W+K\)

\(h \nu=h \nu_{o}+e V_{o}\)

Put the given values in above equation,

\(h \nu=6.2 e V+5 e V\)

\(=11.2 e V\)

We know that,

\(h \nu=\frac{h c}{\lambda}\)

\(\Rightarrow\frac{h c}{\lambda}=11.2 e V\)

\(\Rightarrow c=3×10^{8}m/s\)

\(\Rightarrow\frac{1242 e V~ n m}{\lambda}=11.2 e V\)

\(\Rightarrow\lambda=\frac{1242 e V~ n m}{11.2 e V}\)

\(=110.89~ n m\)

\(\simeq 1100 A^{\circ}\)

Which is under the ultraviolet region.

Hence, the correct option is (B).

Case 1: Stopping potential for \(\lambda_1=V_1\)

\(eV_1=h\nu\Rightarrow\frac{hc}{\lambda_1}\)

\(eV_1=\frac{hc}{\lambda_1}\quad\ldots\)(1)

Case 2: Stoping potential for \(\lambda_2=V_2\)

\(eV_2=h\nu\)

\(eV_2=\frac{hc}{\lambda_2}\)

According to the question,

\(V_2>V_1\)

\(\therefore\frac{hc}{\lambda_2}>\frac{hc}{\lambda_1}\)

\(\Rightarrow \lambda_1>\lambda_2\)

Hence, the correct option is (D).

The de-Broglie wavelength of a particle of mass \(\mathrm{m}\) and moving with velocity \(\mathrm{v}\) is given by,

\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})\)

de-Broglie wavelength of a proton of mass \(\mathrm{m}_{1}\) and kinetic energy \(\mathrm{k}\) is given by,

\(\lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{k}}} \quad(\because \mathrm{p}=\sqrt{2 \mathrm{mk}})\)

\(\Rightarrow \lambda_{1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{qV}}} \ldots . .\) (i) \(\quad[\because \mathrm{k}=\mathrm{qV}]\)

For an alpha particle mass \(\mathrm{m}_{2}\) carrying charge \(\mathrm{q}_{0}\) is accelerated through potential \(\mathrm{V}\), then,

\(\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{2} \mathrm{q}_{0} \mathrm{~V}}}\)

\(\because\) For \(\alpha-\) particle \(\left({ }_{2}^{4} \mathrm{He}\right): \)

\(\mathrm{q}_{0}=2 \mathrm{q}\) and \(\mathrm{m}_{2}=4 \mathrm{~m}_{1}\)

\(\therefore \lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{1} \times 2 \mathrm{q} \times \mathrm{V}}}.....\) (ii)

The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get,

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{1} \mathrm{q} \mathrm{V}}} \times \frac{\sqrt{2 \times \mathrm{m}_{1} \times 4 \times 2 \mathrm{q} \mathrm{V}}}{\mathrm{h}}\)

\(=\frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\)

\(\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=2 \sqrt{2}\)

Hence, the correct option is (A).

Given:

The wavelength of the incident light \(=4000\) Å

\(\mathrm{KE}=2\mathrm{eV}\)

The relation for kinetic energy of a photoelectron is given by \(\mathrm{KE}=\mathrm{h\nu}\) - work function or \(\mathrm{KE}=\mathrm{h} \nu-\mathrm{W}\)

\(\mathrm{KE=\frac{h c}{\lambda}-W}\) or \(\mathrm{W=\frac{h c}{\lambda}-KE}\)

\(\mathrm{W=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10}}-2 e V}\)

\(=4.95 \times 10^{-19} \mathrm{~J}-2 \mathrm{eV}\)

\(=\frac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}}-2 \mathrm{eV}\)

\(=3 \mathrm{eV}-2 \mathrm{eV}\)

\(=1 \mathrm{eV}\)

Hence, the correct option is (B).

Given that electron has a mass \({m}\).

De-Broglie wavelength for an electron will be given as,

\(\lambda_{e}=\frac{h}{p} \quad \ldots\)(i)

where,

\({h}\) is the Planck's constant, and

\(p\) is the linear momentum of electron

Kinetic energy of electron is given by, 

\({E}=\frac{\mathrm{p}^{2}}{2 {~m}}\)

\(\Rightarrow p=\sqrt{2 m E} \quad \ldots\) (ii)

From equation (i) and (ii), we have

\(\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \quad \cdots\)(iii)

Energy of a photon can be given as,

\(E=h v\)

\(\Rightarrow E=\frac{h c}{\lambda_{p}}\)

\(\Rightarrow \lambda_{p}=\frac{h c}{E}\quad \cdots\)(iv)

Thus, \(\lambda_{\mathrm{p}}\) is the de-Broglie wavelength of photon.

Now, dividing equation (iii) by (iv), we get

\( \frac{\lambda_{e}}{\lambda_{p}}=\frac{h}{\sqrt{2 m E}} \cdot \frac{E}{h c}\)

\(\Rightarrow  \frac{\lambda_{e}}{\lambda_{p}}=\frac{1}{c} \cdot \sqrt{\frac{E}{2 m}}\)

Hence, the correct option is (D).

Metal A has higher threshold frequency,

We know that for a glven photosensitlve materlal,

The stopping potential varies linearly with the frequency of the incident radiation.

Also,

Work functlon,

\(\phi_{0}=\mathrm{hv}_{0}\)

Also,

\(\phi_{0}=\mathrm{e} \times\) magnitude of intercept on potential axis.

Here for metal A, \(y\) - intercept is more negative.

Hence, the correct option is (B).

We know that the photoelectric effect of the electron is the emission of the electron due to electromagnetic radiation. This photo electron gets excited and leaves the metal, commonly called photo emission.

Then the energy \(E\) of the photoelectron is given by Planck as:

\(E=h \nu=\frac{h c}{\lambda}\) 

Where, \(h\) = Planck's constant, \(\nu\) = frequency of the incident light, \(c\) = speed of light  \(\lambda\) = wavelength of the incident light

Also, the frequency of the incident light and the kinetic energy is related as,

\(E_{p}=\phi+E_{e}\),

Where \(E_{p}\) = energy of the photon, \(\phi\) = work function,  \(E_{e}\) = energy of the photoelectron.

Given that, \(\phi

Hence, the correct option is (D).

Let \(\mathrm{E_{n}}\) and \(\mathrm{E_{m}}\) be the energies of electron in \(n^{\text {th }}\) and \(m^{\text {th }}\) states.

Then, \(\mathrm{E_{n}-E_{m}=h v_{0}} \ldots(1)\)

In the second case when the atom is moving with a velocity \(\mathrm{v}\). Let \(\mathrm{v^{\prime}}\) be the velocity of atom after emitting the photon. Applying conservation of linear momentum,

\(\mathrm{mv}=\mathrm{mv}^{\prime}+\frac{\mathrm{h\nu}}{\mathrm{c}}\) ( \(\mathrm{m}\) = mass of hydrogen atom)

\(\Rightarrow \mathrm{v}^{\prime}=\left(\mathrm{v}-\frac{\mathrm{h\nu}}{\mathrm{mc}}\right) \ldots(2)\)

Applying conservation of energy

\(\mathrm{E}_{\mathrm{n}}+\frac{1}{2} \mathrm{mv}^{2}=\mathrm{E}_{\mathrm{m}}+\frac{1}{2} \mathrm{mv'}^{2}+\mathrm{h}\nu\)

\(\Rightarrow \mathrm{h} \nu=\left(\mathrm{E}_{\mathrm{n}}-\mathrm{E}_{\mathrm{m}}\right)+\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{v'}^{2}\right)\)

From equation (1) and (2)

\(=\mathrm{h}\nu_{0}+\frac{1}{2} \mathrm{~m}\left[\mathrm{v}^{2}-\left(\mathrm{v}-\frac{\mathrm{h} \nu}{\mathrm{mc}}\right)^{2}\right]\)

\(=\mathrm{h}\nu_{0}+\frac{1}{2} \mathrm{~m}\left[\mathrm{v}^{2}-\mathrm{v}^{2}-\frac{\mathrm{h}^{2} \nu ^{2}}{\mathrm{~m}^{2} \mathrm{c}^{2}}+\frac{2 \mathrm{h\nu v}}{\mathrm{mc}}\right]\)

\(=\mathrm{h} \nu_{0}+\frac{\mathrm{h} \nu \mathrm{v}}{\mathrm{c}}-\frac{\mathrm{h}^{2} \nu^{2}}{2 \mathrm{mc}^{2}}\)

Here the term is \(\frac{\mathrm{h}^{2}\nu^{2}}{2 \mathrm{mc}^{2}}\) is very small. So, can be neglected.

\(\therefore \mathrm{h \nu=h \nu_{0}+\frac{h \nu v}{c}}\)

\(\Rightarrow \nu=\nu_0 +\frac{\nu\mathrm{v}}{c}\)

\(\Rightarrow \nu_{0}=\nu\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)\)

Hence, the correct option is (A).

We know that the energy carried by the photon is given by the following equation:-

\(E=p c\) Which can also be written as follows:-

\(p=\frac{E}{c}\).......(i)

Where \(p\) denotes momentum, \(E\) is energy and \(c\) is speed of light.

One important fact is that the momentum incidence on the surface is equal to the momentum reflected from the surface. But reflected momentum is in the opposite direction.

Therefore, change in momentum, \(\Delta p=p_{i}-p_{r}\)

Where \(p_{i}\) and \(p_{r}\) are incident and reflected momentum respectively.

\(p_{i}=\frac{E}{c} \ldots \ldots \ldots \ldots \ldots\)(ii)

Since, \(p_{i}=p_{r}\).......(iii)

So, \(p_{r}=\frac{-E}{c}\)

\(\left(p_{r}\right.\) is in opposite direction of \(p_{i}\) )

But change in momentum is given as follows:-

\(\Delta p=p_{i}-p_{r}\).......(iv)

Putting value of \((i i)\) and \((i i i)\) in \((i v)\), we have

\(\Rightarrow \Delta p=\frac{E}{c}-\frac{-E}{c} \)

\(\Rightarrow \Delta p=\frac{E}{c}+\frac{E}{c}\)

\(\Rightarrow \Delta p=\frac{2 E}{c}\)

Hence, the correct option is (B).

The Einstein's equation for photoelectric effect is,

\(\mathrm{eV}_{0}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}\)

where,

\(\mathrm{V}_{0}=\) stopping potential

\(\lambda=\) wavelength of incident light

\(\mathrm{W}=\) work function of metal.

\(E=4 \mathrm{NC}^{-1}, ~d=1m\)

\(\mathrm{V}_{0}=\frac{\mathrm{E} }{ \mathrm{d}}=\frac{4 }{ 1}=4\) volt

\(\lambda=200 \mathrm{~nm}=200 \times 10^{-9} \mathrm{~m}\)

Thus, \(\mathrm{W}=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}_{0}\)

\(=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{200 \times 10^{-9}}-\left(1.6 \times 10^{-19}\right) 4\)

\(=3.53 \times 10^{-19} \mathrm{~J}\) (as \(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\))

\(=\frac{3.53 \times 10^{-19} }{1.6 \times 10^{-19} }\)

\(=2.2 \mathrm{eV}\)

Hence, the correct option is (D).

Given:

The wavelength of the incident light \(= 4000\) Å

\(=4000 \times 10^{-10}m\)

and the value of the work function is 2eV.

As we know,

Work function \(\phi=h f-K E\)

Where \(h f\) the incident energy of the light and \(\mathrm{KE}\) is the kinetic energy of the ejected photoelectron.

Therefore, \(\mathrm{KE}=h f-\phi\)

We know that \(f=\frac{c}{\lambda}\)

Thus, \(K E=\frac{h c}{\lambda}-\phi\)........(i)

We know that:

\(c=3 \times 10^{8}\)m/s, \(h=606 \times 10^{-34}\), \(\phi=2\)

Put the given values in equation (i),

\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}}-2\quad\) ( \(1.6 \times 10^{-19}\) is used to get the value in terms of eV)

\(\Rightarrow K E=10.37 \mathrm{eV}\)

Therefore, the value of the kinetic energy of the ejected photoelectron is \(10.37 \mathrm{eV}\).

Hence, the correct option is (D).

Given:

\(\mathrm{\nu}_{1}=3.2 \times 10^{16}Hz\)

\(\mathrm{\nu}_{2}=2 \times 10^{16}Hz\)

\((\mathrm{KE})_{1}=\mathrm{h\nu}_{1}-\mathrm{h\nu}_{0}\).....(i)

\((\mathrm{KE})_{2}=\mathrm{h\nu}_{2}-\mathrm{h\nu}_{0}\).....(ii)

According to the question,

\((\mathrm{KE})_{1}=2 \times(\mathrm{KE})_{2}\).....(iii)

Put the values from (i) and (ii) in (iii),

\( \mathrm{h\nu}_{1}-\mathrm{h\nu}_{0}=2\left(\mathrm{h\nu}_{2}-\mathrm{h\nu}_{0}\right)\)

\(\Rightarrow \mathrm{h\nu}_{0}=2 \mathrm{h\nu}_{2}-\mathrm{h\nu}_{1}\)

\(\Rightarrow \mathrm{\nu}_{0}=2 \mathrm{\nu}_{2}-\mathrm{\nu}_{1}\)

\(=2 \times\left(2 \times 10^{16}\right)-\left(3.2 \times 10^{16}\right)\)

\(=0.8 \times 10^{16} \mathrm{~Hz}\)

\(=8 \times 10^{15} \mathrm{~Hz}\)

Hence, the correct option is (B).

Given:

Two photons of same frequency are produced due to the annhiliation of a proton and antiproton.

We know that:

\(c=3 \times 10^{8}\)

Mass of a proton = \(1.67 \times 10^{-27}\)

\(E=m c^{2}\)

Put the values in above formula.

\(E=\left(2 \times 1.67 \times 10^{-27}\right) \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}\)

\(=3.006 \times 10^{-10} \mathrm{~J}\)

Also We know that:

\(2 h \nu=E\) or \(2 h \frac{c}{\lambda}=E\)

\(\therefore \lambda=\frac{2 h c}{E}\)

Put the values in above formula.

\(=\frac{2 \times 6.62 \times 10^{-34} \times 3 \times 10^{8}}{3.006 \times 10^{-10}} \mathrm{~m}\)

\(=1.323 \times 10^{-15} \mathrm{~m}\)

Hence, the correct option is (B).

For de-Broglie wavelength,

\(\lambda_{1} =\frac{h}{p}=\frac{h}{\sqrt{2 m K}}\)

Where \(m= \) mass and \(K =\) kinetic energy 

According to the question the kinetic energy of the particle is increased to 16 times its previous value,

\(\lambda_{2} =\frac{h}{\sqrt{2 m 16 K}}\)

\(=\frac{h}{4 \sqrt{2 m K}}\)

\(\Rightarrow \lambda _{2}=\frac{\lambda_{1}}{4}\)

The percentage change in the de-Broglie wavelength of the particle is:

\(\Delta \lambda \% = \frac{\lambda _{1}- \lambda _{2}}{\lambda _{1}}×100\)

Put the value of \(\lambda _{2}\) in above formula:

\( \Delta \lambda \% = \frac{\lambda _{1}- \frac{\lambda _{1}}{4}}{\lambda _{1}}×100\)

\(\Rightarrow \Delta \lambda \% = \frac{\frac{3}{4}\lambda _{1}}{\lambda_{1}}×100\)

\(\Rightarrow \Delta\lambda \%=75 \)

Hence, the correct option is (B).

Given:

\(\lambda=600 \mathrm{~nm}\)

We know that

\(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

\(h=4.1357 \times 10^{-15} \mathrm{eV}\)

The energy of the photon with a wavelength lambda is given by \(E=\frac{h c}{\lambda}\)

Where \(h\) is the planck's constant

\(c\) is the speed of light

\(\lambda\) is the wavelength

Put the values in formula given above.

\(E=\frac{4.1357 \times 10^{-15} \times 3 \times 10^{8} }{600 \times 10^{-9}}\)

\(=\frac{4.1357 \times 3}{6}\)

\(=2.06785 \mathrm{~eV}\)

Hence, the correct option is (D).

As we know that the Einstein's photo-electric equation is given by,

\( K_{\max }=h \nu-\phi_{0}.....(1)\)

Also we know that the frequency of a light wave is given by

\( \nu=\frac{c}{\lambda}.....(2)\)

Substituting the value from (2) to (1), we get:

\( K_{\max }=\frac{h c}{\lambda}-\phi_{0}\)

So the work function of the metal is given by 

\( \phi_{0}=\frac{h c}{\lambda}-K_{\max }.....(3)\)

According to the question,

\(\lambda=400 \mathrm{~nm}, K_{\max }=1.68 \mathrm{eV}\), \(h c=1240 \mathrm{eV} . \mathrm{nm}\)

Putting these values in \((3)\) we get,

\( \phi_{0}=\frac{1240}{400}-1.68\)

\(\Rightarrow \phi_{0}=1.42 \mathrm{eV}\)

Thus the work function of the metal comes out to be equal to \(1.42 \mathrm{eV}\).

Hence, the correct option is (B).

Let the work function for the given metal be \(\phi\). When the light of the frequency \(n\) falls on the surface of the metal, the maximum kinetic energy of the ejected photoelectron is given by,

\(K . E_{\max }=n h-\phi\)

\(\frac{1}{2} m v^{2}=n h-\phi \quad \ldots(i)\)

When the light of the frequency 3n falls on the surface of the metal, the maximum kinetic energy of the ejected photoelectron is given by,

\(K . E_{\max }=3 n h-\phi\)

Let the maximum velocity of the electron be \(v\). 

\(\frac{1}{2} m v^{2}=3 n h-\phi \quad \ldots(i i)\)

Substituting \(n h=\frac{1}{2} m v^{2}+\phi\) from equation (i) to equation (ii) 

\(\frac{1}{2} m v^{2}=3\left(\frac{1}{2} m v^{2}+\phi\right)-\phi\)

\(\frac{1}{2} m v^{2}=\frac{3}{2} m v^{2}\)

\(\frac{1}{2} m v^{2}-\frac{3}{2} m v^{2}=0\)

The work function cannot be negative or zero.

Thus,

\(\frac{1}{2} m v^{2}-\frac{3}{2} m v^{2}>0\)

\(\Rightarrow v >v \sqrt{3}\)

Hence, the correct option is (A).

Given:

\(E=5.5 e V\)

\(\phi=4.5 e V\)

Therefore, the kinetic energy will be equal to

\( K . E=E-\phi\)

On substituting the values in above formula, we get

\( K . E=(5.5-4.5) e V\)

\(\Rightarrow K . E=1 e V\)

Wavelength \((\lambda)=\frac{h}{\sqrt{2 M_{e} \times K \cdot E}}\)

\(\lambda\) is the wavelength.

\(h\) is the Planck's constant \(6.6 \times 10^{-34}\).

\(M_{e}\) is the mass of an electron \(9.1 \times 10^{-31}\).

\(K . E\), is the kinetic energy = \(1.16 \times 10^{-19}\).

So on substituting the values in the formula of wavelength, we get 

\( \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1 \times 1.16 \times 10^{-19}}}\)

\(\Rightarrow \lambda=1.24 \times 10^{-9} m\)

Hence, the correct option is (C).

The mean free path of electrons is defined as the distance travelled by the electrons before collision, here mean free path \(\lambda=4 \times 10^{-8} \mathrm{~m}\).

And the energy provided by the electric field is 2eV.

\(\mathrm{V}_{0}=2 \mathrm{eV}\)

Now, electric field \(\mathrm{E}=\frac{V_{0}}{\lambda}\)

Put the values in above formula.

\(\mathrm{E}=\frac{2}{4  \times 10^{-8}}\)

\(=0.5 \times 10^{8}\)

\(=5 \times 10^{7} \mathrm{Vm}^{-1}\)

Hence, the correct option is (D).

Given:

\(\mathrm{W}=5 \mathrm{eV}=5 \times\left(1.6 \times 10^{-19}\right) \mathrm{J}=8.0 \times 10^{-19} \mathrm{~J}\)

Using Einstein's photoelectric equation, \(\mathrm{E}_{1}=\mathrm{KE}+\mathrm{W}\)

\(\mathrm{E}_{1}=\frac{\mathrm{hc} }{ \lambda}\).....(i)

\(h=6.63×10^{-34}\), \(c=3×10^{8}m/s\), \(\lambda = 200×10^{-9}m\)

Put the given values in (i).

\(=\frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right) }{\left(200 \times 10^{-9}\right)}\)

\(=9.945 \times 10^{-19}\)

\(\mathrm{KE}=\mathrm{E}_{1}-\mathrm{W}\).....(ii)

Put the given values in (ii).

\(=9.945 \times 10^{-19}-8.0 \times 10^{-19}\)

\(=1.945 \times 10^{-19} \mathrm{~J}\)

Kinetic energy of the electron \((\mathrm{KE})=\frac{1 }{ 2 \mathrm{mv}^{2}}\)

\(\Rightarrow \mathrm{v}=[\frac{2 \mathrm{KE} }{ \mathrm{m}}]^{\frac{1 }{ 2}} \).....(iii)

Put the given values in (iii).

\(\mathrm{v}=\left[\frac{2\left(1.195 \times 10^{-19}\right) }{ 9.1 \times 10^{-31}}\right]^{\frac{1 }{ 2}} \)

\(=6.54 \times 10^{2} \mathrm{~m} / \mathrm{s}\)

Hence, the correct option is (A).

As, we know de-Broglie wavelength,

\(\lambda=\frac{\mathrm{h}}{\mathrm{p}} \)

\(\therefore \lambda \propto \frac{1}{\mathrm{p}} \)

\(\Rightarrow \frac{\Delta \mathrm{p}}{\mathrm{p}}=-\frac{\Delta \lambda}{\lambda}\)

\(\therefore\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right| \)

Change in the de-Broglie wavelength is found to be 0.20%.

\(\Rightarrow \frac{\Delta\mathrm{p}}{\mathrm{p}}=\frac{0.20}{100}=\frac{1}{500} \)

\(\mathrm{p}=500 \Delta\mathrm{p}\)

Hence, the correct option is (C).

The de-Broglie wavelength is given by

\(\lambda=\frac{h}{P}\)

\(\Rightarrow P \lambda=h\)

This equation is in the form of \(y x=c\), which is the equation of a rectangular hyperbola.

Hence, the correct option is (B).

A laser device produces amplification in the Ultraviolet or visible region.

Laser, a device that stimulates atoms or molecules to emit light at particular wavelengths and amplifies that light, typically producing a very narrow beam of radiation. The emission generally covers an extremely limited range of visible, infrared, or ultraviolet wavelengths.

Hence, the correct option is (B).

Given that when wavelength is \(\lambda\) then the speed of the electron is \(v\). Let us assume that when wavelength is \(\frac{3 \lambda}{4}\) then the speed of electron is \(v_{1}\).

According to the equation of photoelectric emission:

\(\frac{\mathrm{hc}}{\lambda}=\frac{1}{2} \mathrm{mv}^{2}+\phi \quad \ldots . .\)(1)

 When the wavelength \(\frac{3 \lambda}{4}\). Then the speed of electron is \(v_{1}\).

\(\frac{4 \mathrm{hc}}{3 \lambda}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\phi \quad \ldots . .\)(2)

On dividing (2) by (1)

\(\frac{4}{3}=\frac{\frac{1}{2} m v_{1}^{2}+\phi}{\frac{1}{2} m v^{2}+\phi}\)

\(\Rightarrow \frac{1}{2} m v_{1}^{2}=\frac{4}{3}\left(\frac{1}{2} m v^{2}\right)+\frac{4\phi}{3}-\phi\)

\(\Rightarrow \frac{1}{2} m v_{1}^{2}=\frac{4}{3}\left(\frac{1}{2} m v^{2}\right)+\frac{\phi}{3}\)

\(\text { or } v_{1}=\text { greater than } v\left(\frac{4}{3}\right)^{\frac{1 }{ 2}}\)

Hence, the correct option is (D).

Let us assume that the threshold frequency of the sphere is \(\lambda_{o} .\) Let the stopping potential is \(V'\) of the surface when the light of wavelength \(\lambda_{3}\) is used. Thus according to Einstein photoelectric equation.

\(\frac{h c}{\lambda_{1}}=\frac{h c}{\lambda_{o}}+e V\).....(i)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{o}}+3 e V\).....(ii)

\(\frac{h c}{\lambda_{3}}=\frac{h c}{\lambda_{o}}+e V^{\prime}\).....(iii)

From equation (i) and (ii)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+3\left(\frac{h c}{\lambda_{1}}-\frac{h c}{\lambda_{0}}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+\frac{3 h c}{\lambda_{1}}-\frac{3 h c}{\lambda_{0}}\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{0}}\)\(\ldots(i v)\)

From equation (iii) and (iv)

\(\frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-2\left(\frac{h c}{\lambda_{3}}-e V^{\prime}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{3}}+2 e V^{\prime}\)

\(\Rightarrow h c\left(\frac{1}{\lambda_{2}}+\frac{2}{\lambda_{3}}-\frac{3}{\lambda_{1}}\right)=2 e V^{\prime}\)

\(\Rightarrow \frac{h c}{e}\left(\frac{1}{2 \lambda_{2}}+\frac{1}{\lambda_{3}}-\frac{3}{2 \lambda_{1}}\right)=V^{\prime}\)

Hence, the correct option is (C).

Let the equation of wave is given by \(y=A \sin (\omega t-k x)\)

Where, \(A\) is amplitude

\(\omega\) is angular frequency

\(k\) is wave number

Now,

Velocity of wave \((v)=\frac{\omega}{k}\quad\ldots\)(1)

Maxinum velocity of particle \((v_p)=A \omega\quad\ldots\)(2)

Comparing both equation, we get

\(A\omega=\frac{\omega}{k}\)

\(\Rightarrow A=\frac{1}{k}\quad\ldots\)(3)

but, k=\(\frac{2\pi}{\lambda}\)

by substituting value of \(k\) in (3),

\(A=\frac{1}{\frac{2\pi}{\lambda}}\)

\(\Rightarrow A=\frac{\lambda}{2 \pi}\)

Therefore, \(A<\frac{\lambda}{2 \pi}\)

Hence, the correct option is (A).

The Einstein's photoelectric equation is:

\(h v=h v_{o}+K_{\max }\)

So,

\(K_{\max }=h v-h v_{0}\)

\(=h v+\text { constant }\)

On comparing it with

\(y=m x+c\)

This is the equation of straight line having positive slope (h) and negative intercept \(h v_{0}\).

Hence, the correct option is (A).

Given:

\(\phi=2.8 \mathrm{eV}, \mathrm{E}=2 \mathrm{eV}\)

We know that Maximum kinetic energy \((\mathrm{E})=\mathrm{h} \nu-\phi\)

Put the given values in above formula.

\(2=\mathrm{h} \nu-2.8\)

\(\Rightarrow \mathrm{h} \nu=4.8 \mathrm{eV}\)

New frequency \( v^{\prime}=2 \nu\)

So, \(E^{\prime}=h \nu^{\prime}-\phi\)

\(\Rightarrow \mathrm{E}^{\prime}=2 \mathrm{~h} \nu-\phi\)

Put the given values in above formula.

\(=2 \times 4.8-2.8\)

\(=6.8 \mathrm{eV}\)

Hence, the correct option is (A).

Given:

Incident Energy is increased by \(20 \%\).

\(KE_{1}=0.5 \mathrm{eV}=\) initial kinetic energy

\(K E_{2}=0.8 \mathrm{eV}=\) final kinetic energy

\(E_{1}=E=\) initial incident energy

According to the question.

\(E_{2}=E_{1}+E_{1} \times 20 \%\)

\(=E_{1}+0.2 E_{1}\)

\(=1.2 \mathrm{E}_{1}=1.2 \mathrm{E}\)

Let \(\phi\) is the work function of metal. According to Einstein equation:- 

\( E=\phi+K E \)

\(E_{1} =\phi+K E_{1}=E \)

Put the value of \(K E_{1}\) in above equation,

\(E_{1} =\phi+0.5=E \).....(i)

\(E_{2} =\phi+K E_{2}=1.2 E \)

Put the value of \(K E_{2}\) in above equation,

\( E_{2}=\phi+0.8=1.2 E\).....(ii)

Divide equation (i) by equation (ii),

\(\frac{E_{1}}{E_{2}}= \frac{\phi+0.5}{\phi+0.8}=\frac{E}{1.2E}=\frac{1}{1.2}\)

\(\Rightarrow 1.2 \phi +0.6=\phi+0.8 \)

\(\Rightarrow 0.2 \phi =0.2 \)

\(\phi =\frac{0.2}{0.2}=1 \mathrm{eV}\)

Work function of metal is \(\phi=1 \mathrm{ev}\).

Hence, the correct option is (B).