Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 62

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Question 1
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A ray of energy 14.2 Me V is emitted from a $$^{60}Co$$ nucleus. The recoil energy of the co-nucleus is nearly

A
$$3 \times 10^{-10}J$$

B
$$3 \times 10^{-15}J$$

C
$$3 \times 10^{-14}J$$

D
$$3 \times 10^{-13}J$$

Solution

$$P_r = P_{co'}$$

$$ P_r = \dfrac{E}{c}$$,

$$KE = \displaystyle \dfrac{P^2}{2m} = \dfrac{E^2}{2c^2m_{co}}$$

$$=\displaystyle \dfrac{(14 \times 1.6 \times 10^{-13})^2}{2 \times(3\times 10^8 \times 60 \times 1.67 \times 10^{-27}}$$

$$= 3 \times 10^{-10}J$$
So, the answer is option (A).
Question 2
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A proton and an electron are accelerated by the same potential difference. Let $$\displaystyle \lambda _{c}$$ and $$\displaystyle \lambda _{p}$$ denote the de Broglie wavelengths of the electron and the proton respectively.

A
$$\displaystyle \lambda _{c}= \lambda _{p}$$

B
$$\displaystyle \lambda _{c}< \lambda _{p}$$

C
$$\displaystyle \lambda _{c}> \lambda _{p}$$

D
The relation between $$\displaystyle \lambda _{c}$$ and $$\displaystyle \lambda _{p}$$ depends on the accelerating potential difference.

Solution

de Broglie wavelength of a particle of mass m moving under a potential V is given my:

$$\lambda=\dfrac{h}{\sqrt{2meV}}$$.

So it is clear that de Broglie wavelength is inversely proportional to the square root of the mass of the material. Now mass of electron is less that the mass of the proton so de Broglie wavelength of electron will be greater than that of proton.
So, the answer is option (C).
Question 3
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The work functions for three different metals $$A$$, $$B$$ and $$C$$ are $$\varphi_{A}$$ >$$\varphi_{B}$$>$$\varphi_{C}$$. The graphs between stopping potential $$(V_{0})$$ and frequency of v of incident radiation for them would look like

A

B

C

D

Solution

We know that $$eV_0=h\nu-\phi$$ (in photoelectric effect)
or $$V_0=(h/e)\nu-\phi$$ This is similar as $$(y=mx-c)$$
As $$\phi_A>\phi_B>\phi_C$$, so intercept is large for A and small for C.
So, the answer is option (B).
Question 4
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The wave nature of particles was studied using diffraction of particle beams by crystal lattices. The wavelength of the waves associated with fast moving particles was found to be in agreement with the de Broglie relation.
For a particle of mass $$m$$ moving with kinetic energy $$E$$, the de Broglie wavelength is

A
$$\displaystyle \dfrac{h}{2mE}$$

B
$$\displaystyle h\sqrt{2mE}$$

C
$$\displaystyle \dfrac{h}{\sqrt{2mE}}$$

D
$$\displaystyle h\sqrt{\dfrac{2}{mE}}$$

Solution

The energy of photon is
$$E = h\nu$$
According to Einstein's energy mass relation
$$E = mc^2$$
From above equations
$$h\nu = mc^2$$

$$\dfrac{hc}{\lambda} = pc$$ ..................(since, $$p = mc$$)

$$\lambda = \dfrac{h}{p}$$

This is the De-Broglie's equation for wavelength for the photon.
For the particle having mass m and velocity v, De-Broglie's equation for wavelength becomes

$$\lambda = \dfrac{h}{mv}$$ .................(1)

The energy of the particle is

$$E = \dfrac{1}{2}mv^2$$

$$E = \dfrac{p^2}{2m}$$

$$p = \sqrt {2mE}$$
using this value in (1), we get

$$\lambda = \dfrac{h}{\sqrt {2mE}}$$

So, the answer is an option (C).
Question 5
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In an experiment on the photoelectric effect, an evacuated photocell with a pure metal cathode is used. Which graph best represent the variation of V, the minimum potential difference needed to prevent current from flowing, when x, the frequency of the incident light, is varied?

A

B

C

D

Solution

The Einstein's equation for photoelectric effect is given by , $$K_{max}=hf-W$$
or $$eV=hf-hf_0$$ where $$f=$$ frquency of incident photon and $$f_0=$$ threshold frequency.
So, $$V=\dfrac{hf}{e}-\dfrac{hf_0}{e}$$
Thus it will represent a straight line like $$y=mx+c$$ with slope, $$m=h/e$$ and intercept $$=\dfrac{-hf_0}{e}$$.
Hence option B will be the correct curve.
Question 6
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De Broglie wavelength of neutrons in thermal equilibrium is (given $$m_n=1.6\times 10^{-27} kg)$$

A
$$30.8/\sqrt T\,\overset {o}{A}$$

B
$$3.08/\sqrt T\,\overset {o}{A}$$

C
$$0.308/\sqrt T\,\overset {o}{A}$$

D
$$0.0308/\sqrt T\,\overset {o}{A}$$

Solution

The de Broglie wavelength, $$\lambda=\dfrac{h}{\sqrt{2mkT}}$$ where $$h=$$ Planck's constant, $$m=$$ mass of the particle , $$k=$$ Boltzmann's constant, $$T=$$ temperature.

So, $$\lambda=\dfrac{6.62\times 10^{-34}}{\sqrt{2\times (1.6\times 10^{-27})\times (1.38\times 10^{-23})T}}=31.5\times 10^{-10} m/\sqrt T \sim 30.8/\sqrt T A^o$$
Question 7
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Which of the following graphs correctly represents the variation of particle momentum with associated de Broglie wavelength?

A

B

C

D

Solution

From de Broglie hypothesis, the momentum of a particle , $$p=\dfrac{h}{\lambda}$$ where $$h=$$ Plank's constant and $$\lambda=$$ wavelength.
So the momentum versus wavelength graph is like $$y=1/x $$ graph, which is represented in graph D.
Question 8
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An $$\alpha-particle$$ and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The $$\alpha-particle$$ and the proton move such that radius of curvature of their paths is same. Find the ratio of their de Broglie wavelengths :

A
$$2:3$$

B
$$3:4$$

C
$$5:7$$

D
$$1:2$$

Solution

Magnetic force, $$\vec{F_B}=q(\vec{v}\times \vec B)=qvB\sin\theta$$

As magnetic field is perpendicular to velocity so $$\theta=90^o$$ so $$F_B=qvB$$

or, $$\dfrac{mv^2}{r}=qvB$$ or $$mv=qBr$$

So, momentum $$p=mv=qBr$$

de Broglie wavelength for proton , $$\lambda_p=\dfrac{h}{p_{p}}=\dfrac{h}{q_pBr_p}$$ and

de Broglie wavelength for alpha particle , $$\lambda_{\alpha}=\dfrac{h}{p_{\alpha}}=\dfrac{h}{q_{\alpha}Br_{\alpha}}$$

Given, $$\dfrac{r_{\alpha}}{r_p}=1$$ and $$q_{\alpha}=2q_p$$

Thus, $$\dfrac{\lambda_{\alpha}}{\lambda_p}=\dfrac{r_pq_p}{r_{\alpha}q_{\alpha}}=1/2$$
Question 9
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A homogeneous ball $$(mass=m)$$ of ideal black material at rest is illuminated with a radiation having a set of photons $$(wavelength=\lambda)$$, each with the same momentum and the same energy. The rate at which photons fall on the ball is n. The linear acceleration of the ball is:

A
$$m\lambda / nh$$

B
$$nh/m\lambda$$

C
$$nh/(2\pi)(m\lambda)$$

D
$$2pm\lambda / nh$$

Solution

According to de-Broglie relation,

$$ \lambda = \dfrac{h}{p}$$

p = linear momentum = $$ m \times velcoity $$

for 1 photon,

$$ \lambda = \dfrac{h}{mv} $$

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

Since n= no. of photons fall on ball pers sec. and

Linear acceleration of the ball = Rate of change of velocity

So, $$ v = n \times \dfrac{h}{m\lambda}$$
Question 10
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In a photoelectric emission, electrons are ejected from metals X and Y by light of frequency f. The potential difference V required to stop the electrons is measured for various frequencies. If y has a greater work function than X, which graph illustrates the expected results?

A

B

C

D

Solution

The answer is option A.
First, the gradient of the graph cannot change (always = h/e ), so answers are A or D.
If Y has a greater work function than X, then the graph for Y should have a more negative y-intercept. therefore figure in option A depicts the concept.