Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Protons are accelerated from rest by a potential difference 4 kV and strike a metal target. If a proton produces one photon on impact of minimum wavelength $$\lambda_1$$ and similarly an electron accelerated to 4 kV strikes the target and produces a minimum wavelength $$\lambda_2$$ then

A
$$\lambda_1 = \lambda_2$$

B
$$\lambda_1 > \lambda_2$$

C
$$\lambda_1 < \lambda_2$$

D
no such relation can be established

Solution

If electrons are accelerated to a velocity $$v$$ by a potential difference V and then allowed to collide with a metal target, the maximum frequency of the X-rays emitted is given by the equation:

$$\dfrac {1}{2} mv^2= eV = h\nu $$

or, $$\displaystyle \lambda_{min} = \dfrac{1240}{V} $$.

So this shows that the minimum wavelength is inversely proportional to the accelerating voltage. Here accelerating voltage are same so the minimum wavelength is same.
So, the answer is option (A).
Question 2
1 / -0

In the following diagram if $$\displaystyle V_{2}>V_{1}$$then

A
$$\displaystyle \lambda _{1}=\sqrt{\lambda _{2}}$$

B
$$\displaystyle \lambda _{1}<\lambda _{2}$$

C
$$\displaystyle \lambda _{1}=\lambda _{2}$$

D
$$\displaystyle \lambda _{1}>\lambda _{2}$$

Solution

Using Einstein's equation of photoelectric effect:

$$eV_1=\dfrac{hc}{\lambda_2}-W$$ ...(1)

where, $$W=$$ work function of metal

and $$eV_2=\dfrac{hc}{\lambda}_2-W$$ ...(2)

From (1) and (2):

$$eV_2-eV_1=\dfrac{hc}{\lambda_2}-\dfrac{hc}{\lambda_1}$$

$$V_2-V_1=\dfrac{hc}{e}(\dfrac{1}{\lambda_2}-\dfrac{1}{\lambda_1})$$

As $$V_2>V_1 $$ so $$ (1/\lambda_2-1/\lambda_1)>0$$ or $$1/\lambda_2>1/\lambda_1$$ or $$\lambda_1>\lambda_2$$
So, the answer is option (D).
Question 3
1 / -0

The kinetic energy of photoelectrons depends on

A
The sum of threshold frequency and frequency of incident light

B
The ratio of threshold frequency and frequency of incident light

C
The difference of threshold frequency and frequency of incident light

D
The intensity of incident light

Solution

The photoelectric effect related with the incident light frequency $$(\nu)$$ by the following equation:
$$E_k=eV_s=h\nu-\phi$$,
where, $$\phi=h\nu_0$$ is the work function of the material and $$E_k$$ is the kinetic energy of the photo electron.
So, $$E_k=h\nu-h\nu_0$$, $$\nu_0$$ is the threshold frequency for the metal.
or, $$E_k=h(\nu-\nu_0)$$.
So the kinetic energy of photoelectrons depends on the difference of threshold frequency and frequency of incident light.
So, the answer is option (C).
Question 4
1 / -0

An electron is confined in the region of width 1 $$ A^{\circ} $$. Estimate its K.E.

A
3.4 eV

B
3.8 eV

C
13.6 eV

D
10.2 eV

Solution

From Heisenberg's uncertainty relation:
$$\Delta P=h/\Delta x=1.1\times 10^{-24}$$, where $$\Delta x=10^{-10} m$$

$$KE=\dfrac{P^2}{2m}=6.1\times ^{-19} J=3.8 eV$$
Question 5
1 / -0

Neglecting variation of mass with energy the wavelength associated with an electron having a kinetic energy $$E$$ is proportional to

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

B
$$E$$

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

D
$$E^{-2}$$

Solution

$$E=\dfrac{mv^2}{2}$$
momentum p = mv
this gives $$p^2=2Em$$
also $$\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2Em}}$$
Question 6
1 / -0

If $$\lambda_1$$ and $$\lambda_2$$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom, then $$\lambda_1/\lambda_2$$ is equal to :

A
$$2$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{1}{4}$$

D
$$4$$

Solution

The De-Broglie wavelengths can be given by:

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

Since mass of electron remains same, we get

$$\dfrac{\lambda _{1}}{\lambda _{2}}=\dfrac{v_{2}}{v_{1}}..(2)$$

Now, by Bohr's postulates, we know that velocity of electron in $$n^{th}$$ orbit in inversely proportional to $$n$$.

Hence, Answer= $$1/2$$.
Question 7
1 / -0

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

A
$$m_2/m_1$$

B
$$m_1/m_2$$

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

D
1:1

Solution

By momentum conservation, $$Mv=m_1v_1+m_2v_2 ...(1)$$
As M is initially rest so $$v=0$$
Now (1) becomes, $$0=m_1v_1+m_2v_2$$
or $$m_1v_1=-m_2v_2$$, it means that the magnitude of momentum of both masses are equal i.e $$p_1=p_2$$
From de Broglie hypothesis, $$\lambda_1=h/p_1$$ and $$\lambda_2=h/p_2$$
So, $$\dfrac{\lambda_1}{\lambda_2}=\dfrac{p_2}{p_1}=1$$ (as $$p_1=p_2$$)
Question 8
1 / -0

A particle of mass m is projected from ground with velocity u making angle $$\theta$$ with the vertical. The de Broglie wavelength of the particle at the highest point is

A
$$\infty$$

B
$$h/mu sin\theta$$

C
$$h/mu cos\theta$$

D
$$h/mu$$

Solution

De Broglie wavelength is given by: $$\lambda=\dfrac{h}{mv}$$

At the highest point, only the horizontal component of the velocity will be there, which is $$u\sin{\theta}$$.

Thus, wavelength is given by, $$\lambda=\dfrac{h}{mu\sin{\theta}}$$
Question 9
1 / -0

Two electrons are moving with the same velocity, one electron enters a region of uniform electric field while the other enters a region of uniform magnetic field. Then, after sometime, if the de-Broglie wavelength of two are $$\lambda_{1}$$ and $$\lambda_{2}$$ respectively, then

A
$$\lambda_{1} = \lambda_{2}$$

B
$$\lambda_{1} > \lambda_{2}$$

C
$$\lambda_{1} < \lambda_{2}$$

D
$$\lambda_{1}>\lambda_{2}$$ or $$\lambda_{1}< \lambda_{2}$$

Solution

Energy of an electron $$E=\dfrac { hc }{ \lambda } $$
where, $$\lambda $$ is the de-Broglie wavelength of the electron.
Therefore, Energy is inversely proportional to the de-Broglie wavelength of an electron.
After some time, electron that has entered electric field has an increase in velocity along the direction of field, which results in the increase of translational kinetic energy without saturation.
The electron which entered the magnetic field has an increased rotational kinetic energy.
Hence, the electron in electric field will have more energy and low de-Broglie wavelenth.
$$\Rightarrow \lambda_2>\lambda_1$$
So, the answer is option (C).
Question 10
1 / -0

The energy $$E$$ and the momentum $$p$$ of a photon is given by $$E = hv$$ and $$p = \displaystyle\ \frac{h}{\lambda}$$. The velocity of photon will be

A
$$E/p$$

B
$$Ep$$

C
$$(E/P)^{2}$$

D
$$\sqrt{E/P}$$

Solution

$$E=\dfrac{hv}{\lambda}$$
also $$p=\dfrac{h}{\lambda}$$
from above two equation
$$v=\dfrac{E}{p}$$

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Dual Nature of Radiation and Matter Test - 30

Result

Dual Nature of Radiation and Matter Test - 30

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Question 1

$$\lambda_1 = \lambda_2$$

$$\lambda_1 > \lambda_2$$

$$\lambda_1 < \lambda_2$$

no such relation can be established

Solution

Question 2

$$\displaystyle \lambda _{1}=\sqrt{\lambda _{2}}$$

$$\displaystyle \lambda _{1}<\lambda _{2}$$

$$\displaystyle \lambda _{1}=\lambda _{2}$$

$$\displaystyle \lambda _{1}>\lambda _{2}$$

Solution

Question 3

The sum of threshold frequency and frequency of incident light

The ratio of threshold frequency and frequency of incident light

The difference of threshold frequency and frequency of incident light

The intensity of incident light

Solution

Question 4

3.4 eV

3.8 eV

13.6 eV

10.2 eV

Solution

Question 5

$$E^{1/2}$$

$$E$$

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

$$E^{-2}$$

Solution

Question 6

$$2$$

$$\dfrac{1}{2}$$

$$\dfrac{1}{4}$$

$$4$$

Solution

Question 7

$$m_2/m_1$$

$$m_1/m_2$$

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

1:1

Solution

Question 8

$$\infty$$

$$h/mu sin\theta$$

$$h/mu cos\theta$$

$$h/mu$$

Solution

Question 9

$$\lambda_{1} = \lambda_{2}$$

$$\lambda_{1} > \lambda_{2}$$

$$\lambda_{1} < \lambda_{2}$$

$$\lambda_{1}>\lambda_{2}$$ or $$\lambda_{1}< \lambda_{2}$$

Solution

Question 10

$$E/p$$

$$Ep$$

$$(E/P)^{2}$$

$$\sqrt{E/P}$$

Solution

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