Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 31

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Question 1
1 / -0

The radius of the second orbit of an electron in hydrogen atom is $$2.116\overset {o}{A}$$. The de Broglie wavelength associated with this electron in this orbit would be

A
$$6.64\overset {o}{A}$$

B
$$1.058\overset {o}{A}$$

C
$$2.116\overset {o}{A}$$

D
$$13.28\overset {o}{A}$$

Solution

From Bohr's postulate, $$mvr=\dfrac{nh}{2\pi}$$ ... (1) where $$m=$$ mass of electron, $$v=$$ orbital velocity of electrons , $$r=$$ radius of the orbit and $$h=$$ Planck's constant
de Broglie wavelength, $$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$$ ....(2)

From (1) and (2), $$\lambda=\dfrac{2\pi r}{n}=\dfrac{2\times 3.14\times 2.116}{2}=6.64 A^o$$
Question 2
1 / -0

When a high frequency electromagnetic radiation is incident on a metallic surface, electrons are emitted from the surface. Energy of emitted photoelectrons depends only on the frequency of incident electromagnetic radiation and the number of emitted electrons depends only on the intensity of incident light.
Einstein's photoelectric equation $$[K_{max}=h\nu-\phi]$$ correctly explains the PE, where $$\nu=$$ frequency of incident light and $$\phi=$$ work function.
The slope of the graph shown in fig. [here h is the Planck's constant and e is the charge of an electron] is

A
$$\dfrac {h}{e}$$

B
$$eh$$

C
$$h$$

D
$$\dfrac {e}{h}$$

Solution

We have : Using photoelectric Effect
$$E_{gnc}=\phi+KE_{max}.... (1)$$
where $$E_{gnc}=$$ Incident energy
$$=hv$$
$$p=$$ work function
$$KE_{max}=Max.k.Energy$$ of emitted photoelectrons
where $$KE_{max}=eV_{o}$$, where
$$V_{o}$$ is the stopping potential
Hence $$eV_{o}=E_{gnc}-\phi$$
$$\Rightarrow eV_{o}=\dfrac{hc}{\lambda}-\phi$$
$$V_{o}=\dfrac{hc}{\lambda e}-\dfrac{\phi}{e}$$
$$\Rightarrow V_{o}=\dfrac{hv}{e}-\dfrac{\phi}{e}... (2)$$
$$\Rightarrow V_{o}\dfrac{h}{e}\left(\dfrac{c}{\lambda}\right)-\dfrac{\phi}{e}$$
Hence Curve between stopping potential and source frequency represent $$\dfrac{h}{e}$$ as the slope from the equation $$(2)$$
Question 3
1 / -0

A particle of mass 3m at rest decays into two particles of masses m and 2m having non-zero velocities. The ratio of the de Broglie wavelengths of the particles $$(\lambda_1/\lambda_2)$$ is

A
1/2

B
1/4

C
2

D
None of these

Solution

By conservation momentum, $$3mv=mv_1+2mv_2 $$
At mass 3m is at rest so $$v=0$$
Now, $$0=mv_1+2mv_2$$ or $$p_1+2p_2=0$$ where $$p_1, p_2$$ are the momentum of m and 2m.
Thus, $$|p_1|=2|p_2| ...(1)$$

So, $$\lambda_1=\dfrac{h}{|p_1|}$$ and $$\lambda_2=\dfrac{h}{|p_2|}$$

Thus, $$\dfrac{\lambda_1}{\lambda_2}=\dfrac{|p_2|}{|p_1|}=1/2$$ using (1)
Question 4
1 / -0

The de Broglie wavelength of a thermal neutron at $$927^oC$$ is $$\lambda$$. Its wavelength at $$327^oC$$ will be

A
$$\lambda /2$$

B
$$\lambda /\sqrt 2$$

C
$$\lambda \sqrt 2$$

D
$$2\lambda$$

Solution

The relation between de Broglie wavelength and temperature is given by, $$\lambda=\dfrac{h}{\sqrt{2mkT}}$$
where $$k=$$ Boltzmann's constant , $$m=$$ mass, $$T=$$ temperature.
Here all are constant except $$T$$ so $$\lambda \propto \dfrac{1}{\sqrt T}$$

Thus, $$\dfrac{\lambda_2}{\lambda_1}=\sqrt{\dfrac{T_1}{T_2}}=\sqrt{\dfrac{273+927}{273+327}}=\sqrt 2$$

or $$\lambda_2=\lambda \sqrt 2$$
Question 5
1 / -0

The energy of a photon is equal to the kinetic energy of a proton. The energy of photon is $$E$$. Let $$\lambda_1$$ be the de Broglie wavelength of the proton and $$\lambda_2$$ be the wavelength of the photon. Then, $$\lambda_1/\lambda_2$$ is proportional to :

A
$$E^0$$

B
$$E^{1/2}$$

C
$$E^{-1}$$

D
$$E^{-2}$$

Solution

The energy photon , $$E=h\nu_2=\dfrac{hc}{\lambda_2}$$ or $$\lambda_2=\dfrac{hc}{E}$$

As energy of photon $$=$$ energy of proton so $$E=\dfrac{p^2}{2m}$$ where p be the momentum of proton.

Now, $$\lambda_1=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mE}}$$

Thus, $$\dfrac{\lambda_1}{\lambda_2}=\dfrac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}}=\dfrac{1}{c\sqrt{2m}}\sqrt E$$

Hence, $$\dfrac{\lambda_1}{\lambda_2}\propto E^{1/2}$$
Question 6
1 / -0

Einstein's Photo electric equation:

A
$$hv =\dfrac{1}{2} mv^2+W $$

B
$$hv =\dfrac{1}{2} mv^2-W $$

C
$$hv =\dfrac{1}{2} mv+W $$

D
$$hv =\dfrac{1}{2} mv-W $$

Solution

Acoording to Einstein's photoelectric equation,

$$hv = \dfrac{1}{2}mv^{2} + \omega \left ( hv_{0} = \omega = workfunction\right ) $$

$$\Rightarrow E = \dfrac{1}{2}mv^{2} = hv - hv_{0}$$
Question 7
1 / -0

We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used the energy required should be

A
1.5 keV

B
15 keV

C
150 keV

D
1.5 MeV

Solution

De broglie wavelength is given by- $$\lambda=\dfrac{h}{mv}$$

$$\implies v=\dfrac{h}{m\lambda}$$

$$\lambda=10pm=10^{-11}m$$

Energy required=$$\dfrac{1}{2}mv^2=\dfrac{1}{e}\dfrac{1}{2}m(\dfrac{h}{m\lambda})^2eV$$

$$=15keV$$
Question 8
1 / -0

Calculate the de Broglie wavelength of an electron having kinetic energy of $$1.6\times 10^{-6}erg$$ $$(m_e=9.11\times 10^{-28}g, h=6.62\times 10^{-27}erg-sec)$$.

A
$$90A^o$$

B
$$0.00029A^o$$

C
$$0.0122A^o$$

D
$$1.29A^o$$

Solution

The expression for the kinetic energy E is $$\displaystyle E = \frac {1}{2} mu^2$$
Hence, $$\displaystyle u = \sqrt {2E}{m}$$......(1)
The De-broglie equation is $$\displaystyle \lambda = \frac {h}{mu}$$.....(2)
Substittue (2) in (1)
$$\displaystyle \lambda = \frac {h}{\sqrt {2mE}}$$
Substitute values in the above expression.
$$\displaystyle \lambda = \frac {6.62 \times 10^{-27}}{\sqrt {2 \times 9.11 \times 10^{-28} \times 1.6 \times 10^{-6}}} = 1.226 \times 10^{-10} cm = 0.01226 A^o$$
Hence, the de Broglie wavelength of the electron is $$\displaystyle 0.01226A^o$$
Question 9
1 / -0

The wavelength of a wave is $$\lambda$$ = 6000 $$\overset{o}{A}$$, then wave number will be

A
1.66 $$\times$$ 10$$^{7}$$ m$$^{-1}$$

B
1.66 $$\times$$ 10$$^{6}$$ m$$^{-1}$$

C
16.6 $$\times$$ 10$$^{-1}$$ m$$^{-1}$$

D
166 $$\times$$ 10$$^{3}$$ m$$^{-1}$$

Solution

Wave number, $$\displaystyle\bar{v}$$ = $$\displaystyle\dfrac{1}{\lambda}$$

$$ = \dfrac{1}{6 \times 10^{-7}} $$

$$= 1.66 \times 10^{6} m^{-1}$$
Question 10
1 / -0

Einstein's photoelectric equation state that
$$hv=W_0 + E_k$$
In the equation, $$E_k$$ refers to the

A
Kinetic energy of all emitted electrons

B
Mean kinetic energy of the emitted electrons

C
Maximum kinetic energy of the emitted electrons

D
Minimum kinetic energy of the emitted electrons

Solution

The photons of a light beam have a characteristic energy proportional to the frequency of the light. In the photoemission process, if an electron within some material absorbs the energy of one photon and acquires more energy than the work function (the electron binding energy) of the material, it is ejected. If the photon energy is too low, the electron is unable to escape the material.
If the photon energy is absorbed, some of the energy liberates the electron from the atom, and the rest contributes to the electron's kinetic energy as a free particle.
Photoelectric effect
$$KE \le hv - W_o$$
$$KE_{max} = hv - W_o$$
The maximum $$KE$$ of electron is when the equality holds.

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Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 31

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Dual Nature of Radiation and Matter Test - 31

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Question 1

$$6.64\overset {o}{A}$$

$$1.058\overset {o}{A}$$

$$2.116\overset {o}{A}$$

$$13.28\overset {o}{A}$$

Solution

Question 2

$$\dfrac {h}{e}$$

$$eh$$

$$h$$

$$\dfrac {e}{h}$$

Solution

Question 3

1/2

1/4

2

None of these

Solution

Question 4

$$\lambda /2$$

$$\lambda /\sqrt 2$$

$$\lambda \sqrt 2$$

$$2\lambda$$

Solution

Question 5

$$E^0$$

$$E^{1/2}$$

$$E^{-1}$$

$$E^{-2}$$

Solution

Question 6

$$hv =\dfrac{1}{2} mv^2+W $$

$$hv =\dfrac{1}{2} mv^2-W $$

$$hv =\dfrac{1}{2} mv+W $$

$$hv =\dfrac{1}{2} mv-W $$

Solution

Question 7

1.5 keV

15 keV

150 keV

1.5 MeV

Solution

Question 8

$$90A^o$$

$$0.00029A^o$$

$$0.0122A^o$$

$$1.29A^o$$

Solution

Question 9

1.66 $$\times$$ 10$$^{7}$$ m$$^{-1}$$

1.66 $$\times$$ 10$$^{6}$$ m$$^{-1}$$

16.6 $$\times$$ 10$$^{-1}$$ m$$^{-1}$$

166 $$\times$$ 10$$^{3}$$ m$$^{-1}$$

Solution

Question 10

Kinetic energy of all emitted electrons

Mean kinetic energy of the emitted electrons

Maximum kinetic energy of the emitted electrons

Minimum kinetic energy of the emitted electrons

Solution

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