Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

If the stationery proton and $$\alpha-$$ particle are accelerated through same potential difference, the ratio of de-Broglie wavelength will be:

A
$$2$$

B
$$1$$

C
$$2\surd{2}$$

D
None of these

Solution

The gain in $$K.E$$. of a charge particle after moving through a potential difference of $$V$$ is given as $$eV$$, that is also equal to $$\dfrac{1}{2}\ mv^{2}$$ where $$v$$ is the velocity of the charge particle.
$$\dfrac{1}{2}\ mv^{2}=qV \quad \Rightarrow \quad v=\sqrt{\dfrac{2qV}{m}}$$
$$\therefore \quad mv=\sqrt{2mqV}$$
$$\Rightarrow \quad$$ de-Broglie wavelength $$=\lambda=\dfrac{h}{mv}=\dfrac{h}{\sqrt{2\ mqV}}$$
$$\therefore \quad \dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha} q_{\alpha} V_{\alpha}}{m_{p} q_{p} V_{p}}}$$
Putting $$V_{\alpha}=V_{p},\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{(4)(2)}{(1)(1)}}=2\sqrt{2}$$
Hence ($$C$$) is correct.
Question 2
1 / -0

The de-Broglie wavelength of an electron in first orbit of Bohr hydrogen is equal to

A
Radius of the orbit

B
Perimeter of the orbit

C
Diameter of the orbit

D
Half of the perimeter of the orbit

Solution

The de-Broglie wavelength $$\lambda=\dfrac{h}{mv}$$ . . . . .($$1$$)
Where the linear momentum of an orbiting electron is given as
$$mvr=\dfrac{nh}{2\pi}$$ for first orbit $$n=1$$
$$\Rightarrow \quad mv=\dfrac{h}{2\pi r}$$ . . . . . ($$2$$)
Using ($$1$$) and ($$2$$) we obtain
$$\lambda =\dfrac { h }{ \left( \dfrac { h }{ 2\pi r } \right) } =2\pi r$$ where, $$2\pi r=$$ perimeter of the orbit
Hence ($$B$$) is correct.
Question 3
1 / -0

If $$E_{1},E_{2}$$ and $$E_{3}$$ represent respectively the kinetic energies of an electron, an $$\alpha-$$particle and a proton each having same de-Broglies wavelength then

A
$$E_{1} > E_{2} > E_{3}$$

B
$$E_{2} > E_{3} > E_{1}$$

C
$$E_{1} > E_{3} > E_{2}$$

D
$$E_{1} = E_{2} = E_{3}$$

Solution
Question 4
1 / -0

If the kinetic energy of a free electron doubles, its de Broglie wavelength changes by the factor

A
$$\cfrac{1}{2}$$

B
$$2$$

C
$$\cfrac{1}{\sqrt{2}}$$

D
$$\sqrt {2}$$

Solution

$$\lambda =\frac { h }{ mv } $$ and  $$k=\dfrac { 1 }{ 2 } { mv }^{ 2 }=\dfrac { { \left( mv \right) }^{ 2 } }{ 2m } $$
$$\Rightarrow mv=\sqrt { 2mk } $$
So,  $$\lambda =\dfrac { h }{ \sqrt { 2mk } } $$
$$\Rightarrow \lambda \sim \dfrac { 1 }{ \sqrt { k } } $$
$$\dfrac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 } } =\dfrac { \sqrt { { k }_{ 1 } } }{ \sqrt { { k }_{ 2 } } } =\dfrac { \sqrt { { k }_{ 1 } } }{ \sqrt { { 2k }_{ 1 } } } $$   $$\left( { k }_{ 2 }=2{ k }_{ 1 } \right) $$
$$\Rightarrow \dfrac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 } } =\dfrac { 1 }{ \sqrt { 2 } } $$
Question 5
1 / -0

After absorbing a slowly moving neutron of mass $${m}_{N}$$ (momentum $$\sim 0$$) a nucleus of mass $$M$$ breaks into two nuclei of masses $${m}_{1}$$ and $$5{ m }_{ 1 }\left( 6{ m }_{ 1 }=M+{ m }_{ N } \right) $$ respectively. If the de-Broglie wavelength of the nucleus with mass $${m}_{1}$$ is $$\lambda$$, the de-Broglie wavelength of the other nucleus will be:

A
$$\lambda$$

B
$$25\lambda$$

C
$$5\lambda$$

D
$$\lambda /5$$

Solution

Momentum should be conserved.
As initial momentum was $$zero $$ so final should also be zero.

so $$p_1 +p_2=0$$ or $$p_1=-p_2$$ i.e. $$both $$ have $$same $$ values..............(1)

where $$p_1$$ is momentum of $$m_1$$ and $$p_2$$ is that of $$m_2$$
The deBroglie wavelength is given by $$\lambda=\dfrac{h}{p}$$
As both mass have same $$momentum, $$ , $$p_1=-p_2$$
so same values of $$wavelength.$$ So answer is $$\lambda$$
Question 6
1 / -0

Photon of frequency $$v$$ has a momentum associated with it. If $$c$$ is the velocity of light, the momentum is :

A
$$v/c$$

B
$$hvc$$

C
$$hv/{ c }^{ 2 }$$

D
$$hv/c$$

Solution

We know,
$$P=\dfrac { h }{ \lambda } $$ where $$\lambda =$$ wavelength
$$P=\dfrac { hv }{ C } $$
Thus, the momentum with $$C$$ being the velocity of light $$=\dfrac { hv }{ C } $$.
Question 7
1 / -0

The graph shown below depicts plot of photocurrent versus anode potential for a cathode with 4 eV work function. The energy of the incident photon is

A
6 eV

B
4 eV

C
2 eV

D
8 eV

Solution

It is given that

Work function

$$ \phi =4\,eV $$

$$ V=2V $$

From Einstein’s relation

$$h\nu =\phi +{{E}_{k}}$$

Where $$h\nu $$ is the energy of photons. And E_k is the kinetic energy

$$ h\nu =(4+2)eV $$

$$ h\nu =6\,eV $$
Question 8
1 / -0

Let $$p$$ and $$E$$ denote the linear momentum and the energy of a photon. For another photon of smaller wavelength (in same medium)

A
Both $$p$$ and $$E$$ increase

B
$$p$$ increases and $$E$$ decreases

C
$$p$$ decreases and $$E$$ increases

D
Both $$p$$ and $$E$$ decrease

Solution

De-Broglie relation:
$$p=\frac { h }{ \lambda } $$
and from Planks relation:
$$E=\frac { hc }{ \lambda } $$
Therefore, as $$\lambda$$ increases, both p and E increase.
Question 9
1 / -0

A proton and electron are accelerated by same potential difference starting from the rest have de-Broglie wavelength $$\lambda_p$$ and $$\lambda_e$$.

A
$$\lambda_e = \lambda_p$$

B
$$\lambda_e < \lambda_p$$

C
$$\lambda_e > \lambda_p$$

D
none of these

Solution

Let $${ m }_{ e }$$ and $${ m }_{ p }$$ be the mass of electron and proton respectively. Let the applied potential different be $$V$$.
Thus, the de-Broglie wavelength of the $${ e }^{ - }$$,
$${ \lambda }_{ e }=\dfrac { h }{ \sqrt { 2{ m }_{ e }eV } } \quad \longrightarrow \left( 1 \right) $$
and de - Broglie wavelength of proton
$${ \lambda }_{ p }=\dfrac { h }{ \sqrt { 2{ m }_{ p }eV } } \quad \longrightarrow \left( 2 \right) $$
Dividing,
$$\dfrac { { \lambda }_{ p } }{ { \lambda }_{ e } } =\dfrac { \sqrt { { m }_{ e } } }{ \sqrt { { m }_{ p } } } $$
$${ m }_{ e }<{ m }_{ p }$$
$$\therefore$$ $$\dfrac { { \lambda }_{ p } }{ { \lambda }_{ e } } <1$$
$$\Rightarrow { \lambda }_{ p }{ \lambda }_{ e }$$
Question 10
1 / -0

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let $${ \lambda }_{ 1 }$$ be de-Broglie wavelength of the proton and $${ \lambda }_{ 2 }$$ be the wavelength of the photon.The ration $$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } }$$ is proportional to:

A
$${ E }^{ 0 }$$

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

C
$${ E }^{ -1 }$$

D
$${ E }^{ -2 }$$

Solution

$$\lambda =\dfrac { h }{ p } $$
photon energy $$E=hv$$
So, $${ \lambda }_{ 1 }=\frac { h }{ p } =\frac { h }{ \sqrt { 2mE } } \quad \longrightarrow \left( i \right) \quad \Rightarrow E=h{ v }_{ 2 }=\dfrac { { h }_{ c } }{ { \lambda }_{ 2 } } $$
$${ \lambda }_{ 2 }=\dfrac { { h }_{ c } }{ E } \quad \longrightarrow \left( ii \right) $$ from eq $$(i)$$ & $$(ii)$$
$$\dfrac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\dfrac { h\sqrt { 2mE } }{ \left( \dfrac { { h }_{ c } }{ E } \right) } \sim \dfrac { E }{ \sqrt { E } } ={ E }^{ 1/2 }$$

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Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 44

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Dual Nature of Radiation and Matter Test - 44

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SHARING IS CARING

Question 1

$$2$$

$$1$$

$$2\surd{2}$$

None of these

Solution

Question 2

Radius of the orbit

Perimeter of the orbit

Diameter of the orbit

Half of the perimeter of the orbit

Solution

Question 3

$$E_{1} > E_{2} > E_{3}$$

$$E_{2} > E_{3} > E_{1}$$

$$E_{1} > E_{3} > E_{2}$$

$$E_{1} = E_{2} = E_{3}$$

Solution

Question 4

$$\cfrac{1}{2}$$

$$2$$

$$\cfrac{1}{\sqrt{2}}$$

$$\sqrt {2}$$

Solution

Question 5

$$\lambda$$

$$25\lambda$$

$$5\lambda$$

$$\lambda /5$$

Solution

Question 6

$$v/c$$

$$hvc$$

$$hv/{ c }^{ 2 }$$

$$hv/c$$

Solution

Question 7

6 eV

4 eV

2 eV

8 eV

Solution

Question 8

Both $$p$$ and $$E$$ increase

$$p$$ increases and $$E$$ decreases

$$p$$ decreases and $$E$$ increases

Both $$p$$ and $$E$$ decrease

Solution

Question 9

$$\lambda_e = \lambda_p$$

$$\lambda_e < \lambda_p$$

$$\lambda_e > \lambda_p$$

none of these

Solution

Question 10

$${ E }^{ 0 }$$

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

$${ E }^{ -1 }$$

$${ E }^{ -2 }$$

Solution

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