Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The figure shows the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. The ratio h/e is

A
$$10^{-15} V s^{-15}V s$$

B
$$2\times 10^{-15}Vs$$

C
$$3\times 10^{-15}Vs$$

D
$$4.14\times 10^{-15}Vs$$

Solution

The Einstein's equation for photoelectric effect is given by $$K=hf-w$$
or $$eV_s=hf-w$$ where $$h=$$ Planck's constant , $$w=$$ work function for metal , $$f=$$ frequency of incident photon.
So, $$V_s=(h/e)f-w/e $$
It will represent a straight line with slope $$m=h/e$$.

From the given plot, $$m=h/e= \dfrac{1.656}{4\times 10^{14}}=4.14\times 10^{-15} V-s$$
Question 2
1 / -0

Representing the stopping potential V along y-axis and $$(1/\lambda)$$ along x-axis for a given photocathode, the curve is a straight line, the slope of which is equal to

A
$$e/hc$$

B
$$hc/e$$

C
$$ec/h$$

D
$$he/c$$

Solution

The Einstein's equation for photoelectric effect is given by , $$K_{max}=h\nu-W$$

or $$eV=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0}$$ where $$\lambda=$$ wavelength of absorbed photon and $$\lambda_0=$$ threshold wavelength.

So, $$V=\dfrac{hc}{e\lambda}-\dfrac{hc}{e\lambda_0}$$

Comparing this equation with general equation of straight line $$y=mx+c$$, we get slope, $$m=hc/e$$
Question 3
1 / -0

An electron of mass $$m_e$$ and a proton of mass $$m_p$$ are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an electron to that associated with proton is

A
1

B
$$m_p /m_e$$

C
$$m_e/m_p$$

D
$$\sqrt {m_p/m_e}$$

Solution

The relation between de Broglie wavelength $$(\lambda)$$ and potential difference $$(V)$$ is given by, $$\lambda=\dfrac{h}{\sqrt{2meV}}$$
As $$V$$ is constant so $$\lambda \propto \dfrac{1}{\sqrt{m}}$$
Thus, $$\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}$$
Question 4
1 / -0

The kinetic energy of a particle is equal to the energy of a photon. The particle moves at 5% of the speed of light. The ratio of the photon wavelength to the de Broglie wavelength of the particle is
[No need to used relativistic formula for the particle.]

A
$$40$$

B
$$4$$

C
$$2$$

D
$$80$$

Solution
Question 5
1 / -0

Directions For Questions

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.

...view full instructions

For photoelectric effect in a metal, the graph of the stopping potential $$V_0$$ (in volt) versus frequency $$v$$ (in hertz) of the incident radiation is shown in Fig. The work function of the metal (in $$eV$$) is

A
12.5

B
14.5

C
16.5

D
18.5

Solution

The x-intercept$$({v}_{0})$$ gives the frequency below which no photoelectric effect will take place.
Hence, work function is given by:
$$W=h{v}_{0}=6.63\times{10}^{-34}\times4\times{10}^{15}=26.52\times{10}^{-19}J$$
Converting to eV, we get:
$$W=16.5eV$$
Question 6
1 / -0

A photon has same wavelength as the de Broglie wavelength of electrons. Given $$C=$$ speed of light, $$v=$$ speed of electron. Which of the following relation is correct? [Here $$E_e=$$ kinetic energy of electron, $$E_{ph}=$$ energy of photon, $$P_e=$$ momentum of electron and $$P_{ph}=$$ momentum of photon]

A
$$E_e/E_{ph}=2C/v$$

B
$$E_e/E_{ph}=v/2C$$

C
$$P_e/P_{ph}=2C/v$$

D
$$P_e/P_{ph}=C/v$$

Solution

Given, $$\lambda_{ph}=\lambda_e ....(1)$$

Now, $$\lambda_e=\dfrac{h}{P_e}=\dfrac{h}{mv} ....(2)$$

Energy of photon, $$E_{ph}=\dfrac{hC}{\lambda_{ph}}=\dfrac{hC}{\lambda_e}$$ (using (1))

Energy of electron, $$E_e=\dfrac{P_e^2}{2m}=\dfrac{h^2}{\lambda_e^2(2m)}$$ using (2)

Now, $$\dfrac{E_e}{E_{ph}}=\dfrac{h^2}{\lambda_e^2(2m)}\times \dfrac{\lambda_e}{hC}=(1/2mC)(h/\lambda_e)=(1/2mC)(mv)=v/2C$$ using (2)

Now, $$\dfrac{P_e}{P_{ph}}=\dfrac{h/\lambda_e}{h/\lambda_{ph}}=1$$ using (1)
Question 7
1 / -0

A particle of mass M at rest decays into two particles of masses $$m_1$$ and $$m_2$$, having non-zero velocities. The ratio of the de Broglie wavelengths of the particles $$\lambda_1/\lambda_2$$ is :

A
$$m_1/m_2$$

B
$$m_2/m_1$$

C
$$1$$

D
$$\sqrt {m_2}/\sqrt {m_1}$$

Solution

By conservation of momentum, $$Mv=m_1v_1+m_2v_2$$

As mass M is at rest so $$v=0$$. Thus, $$m_1v_1=-m_2v_2$$, it indicates the magnitude of momentum of masses is same i.e $$p_1=p_2$$

So, de Broglie wavelengths for masses $$m_1$$ and $$m_2$$ are $$\lambda_1=h/p_1$$ and $$\lambda_2=h/p_2$$

Thus, $$\dfrac{\lambda_1}{\lambda_2}=\dfrac{p_2}{p_1}=1$$ as $$(p_1=p_2)$$
Question 8
1 / -0

Two identical non-relativistic particles A and B move at right angles to each other, processing de Broglie wavelengths $$\lambda_1$$ and $$\lambda_2$$ respectively. The de Broglie wavelength of each particle in their centre of mass frame of reference is :

A
$$\lambda_1+\lambda_2$$

B
$$2\lambda_1\lambda_2/(\sqrt {\lambda_1^2+\lambda_2^2})$$

C
$$\lambda_1\lambda_2/(\sqrt {|\lambda_1^2+\lambda_2^2|})$$

D
$$(\lambda_1+\lambda_2)/2$$

Solution

Two identical non-relativistic particles A and B move at right angles to each other with say velocity $$v_1, v_2$$. along X and Y directions respectively (say).Therefore Velocity of center of mass (COM) ; $$v_c= v_1\hat i+v_2\hat j$$Velocity of 1st particle with respect to COM; $$v_{1c}=v_1-v_c=\dfrac{v_1\hat i-v_2 \hat j}{2}$$
Velocity of 2nd particle with respect to COM; $$v_{2c}=v_2-v_c=\dfrac{v_2\hat j-v_1 \hat i}{2}$$De Broglie wavelength of 1st particle w.r.t COM
$$\lambda_{1c}=\dfrac{h}{mv_{1c}}=\dfrac{2h}{m\sqrt{v_1^2+v_2^2}}$$$$\lambda_{1c}=\dfrac{2h}{m\sqrt{(\dfrac{h}{m\lambda_1})^2+(\dfrac{h}{m\lambda_2})^2}}$$$$\lambda_{1c}=\dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$$Similarly,De Broglie wavelength of 2nd particle w.r.t COM$$\lambda_{2c}=\dfrac{h}{mv_{2c}}=\dfrac{2h}{m\sqrt{v_1^2+v_2^2}}$$$$\lambda_{2c}=\dfrac{2h}{m\sqrt{(\dfrac{h}{m\lambda_1})^2+(\dfrac{h}{m\lambda_2})^2}}$$$$\lambda_{2c}=\dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$$
Question 9
1 / -0

Two electrons are moving with same speed $$v$$. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, when after some time de Broglie wavelengths of two are $$\lambda_1$$ and $$\lambda_2$$, respectively. Now :

A
$$\lambda_1=\lambda_2$$

B
$$\lambda_1 > \lambda_2$$

C
$$\lambda_1 < \lambda_2$$

D
$$\lambda_1$$ can be greater than or less than $$\lambda_2$$

Solution

In a uniform electric field, an electron has a motion with uniform acceleration in the direction of the field with a magnitude which equals $$\dfrac{eE}{m_e}$$ whereas in a uniform magnetic field, an electron undergoes a circular motion with the centripetal force of magnitude $$evB$$. Since the motion is circular in the magnetic field the magnitude of the velocity of the electron will be equal to the initial velocity of $$v$$.

$$\therefore$$ $$\lambda_2 = \dfrac{h}{mv}$$

But in the case of the uniform electric field, the final velocity depends on the direction of the electric field and the time period for which it is applied. If the applied field is opposite to the initial direction then the final velocity could be zero which means $$\lambda_1 > \lambda_2$$ whereas if the applied field is in the direction of initial motion then $$\lambda_1 < \lambda_2$$ after some time.
Question 10
1 / -0

The magnitude of the de-Broglie wavelength $$(\lambda)$$ of electron (e), proton (p), neutron (n) and $$\alpha$$-particle $$(\alpha)$$ all having the same energy of 1MeV, in the increasing order will follow the sequence

A
$$\lambda _e, \lambda _p, \lambda _n, \lambda _\alpha $$

B
$$\lambda _e, \lambda _n, \lambda _p, \lambda _\alpha $$

C
$$\lambda _\alpha , \lambda _n, \lambda _p, \lambda _e $$

D
$$\lambda _p , \lambda _e, \lambda _\alpha , \lambda _n $$

Solution

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

$$p=mv$$

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

$$ p = \sqrt{2mE}$$
De-Broglie wavelength

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

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

Given Energy is same for all
$$ \lambda \sqrt{m} = const.$$

$$ \lambda \propto \cfrac{1}{\sqrt{m}}$$

Hence, lower the mass, higher the wavelength
$$m_{\alpha} > m_{n}> m_p > m_e$$
$$\lambda_{\alpha} < \lambda_{n}< \lambda_p <\lambda_e$$

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Dual Nature of Radiation and Matter Test - 63

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Dual Nature of Radiation and Matter Test - 63

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Question 1

$$10^{-15} V s^{-15}V s$$

$$2\times 10^{-15}Vs$$

$$3\times 10^{-15}Vs$$

$$4.14\times 10^{-15}Vs$$

Solution

Question 2

$$e/hc$$

$$hc/e$$

$$ec/h$$

$$he/c$$

Solution

Question 3

1

$$m_p /m_e$$

$$m_e/m_p$$

$$\sqrt {m_p/m_e}$$

Solution

Question 4

$$40$$

$$4$$

$$2$$

$$80$$

Solution

Question 5

12.5

14.5

16.5

18.5

Solution

Question 6

$$E_e/E_{ph}=2C/v$$

$$E_e/E_{ph}=v/2C$$

$$P_e/P_{ph}=2C/v$$

$$P_e/P_{ph}=C/v$$

Solution

Question 7

$$m_1/m_2$$

$$m_2/m_1$$

$$1$$

$$\sqrt {m_2}/\sqrt {m_1}$$

Solution

Question 8

$$\lambda_1+\lambda_2$$

$$2\lambda_1\lambda_2/(\sqrt {\lambda_1^2+\lambda_2^2})$$

$$\lambda_1\lambda_2/(\sqrt {|\lambda_1^2+\lambda_2^2|})$$

$$(\lambda_1+\lambda_2)/2$$

Solution

Question 9

$$\lambda_1=\lambda_2$$

$$\lambda_1 > \lambda_2$$

$$\lambda_1 < \lambda_2$$

$$\lambda_1$$ can be greater than or less than $$\lambda_2$$

Solution

Question 10

$$\lambda _e, \lambda _p, \lambda _n, \lambda _\alpha $$

$$\lambda _e, \lambda _n, \lambda _p, \lambda _\alpha $$

$$\lambda _\alpha , \lambda _n, \lambda _p, \lambda _e $$

$$\lambda _p , \lambda _e, \lambda _\alpha , \lambda _n $$

Solution

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