Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

In a photocell circuit the stopping potential, $$v_0$$ , is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency v of incident light.
The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $$ e= 1.6 \times 10^{-19} C $$ )

A
$$ 6.4 \times 10^{-34} Js, 2.0$$ eV

B
$$ 6.0 \times 10^{-34} Js, 2.0$$ eV

C
$$ 6.4 \times 10^{-34} Js, 3.2 $$ eV

D
$$ 6.0 \times 10^{-34} Js, 3.2 $$ eV

Solution

Observing the graph and relating it to the Einstein photoelectric equation, we have the graph between stopping potential and frequency as shown in the figure.

The slope of the graph shown is $$\frac{4-(-2)}{(1.6-0)\times 10^{15}}$$
Thus, $$\frac{h}{e}=\frac{6}{1.6\times 10^{15}} \Rightarrow h = 6\times 10^{-34} Js$$

Work function is given by the negative value of Y-intercept.

Y-intercept in the figure is $$-2V \Rightarrow w=2eV$$
Question 2
1 / -0

A proton and an alpha particle are subjected to same potential difference $$V$$. Their de-Broglies wavelengths $$\lambda_{p}, \lambda_{\alpha}$$ will be in the ratio.

A
$$2 : 1$$

B
$$2\sqrt {2} : 1$$

C
$$4 : 1$$

D
$$1 : 2$$

Solution

Let the mass of alpha be $$4m$$ and charge be $$2e$$
therefore mass of proton is $$m$$ and charge is $$e$$
Energy of proton $$E_P=e\times V$$
So momentum is $$P_P=\sqrt{2Em}=\sqrt{2eVm}$$
Similar for alpha $$P_{\alpha}=\sqrt{2\times2e\times V\times 4m}$$
$$\lambda_{p}=\dfrac{h}{P_p}$$
$$\lambda_{\alpha}=\dfrac{h}{P_{\alpha}}$$
$$\dfrac{\lambda_{\alpha}}{\lambda_{P}}=2\sqrt{2}$$
Question 3
1 / -0

A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

A
$$2\sqrt {2}$$

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

C
$$2$$

D
$$\sqrt {2}$$

Solution

Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity $$v$$ is given by

$$\lambda = \dfrac {h}{mv} $$ $$ (\because p = mv)$$

de-Broglie wavelength of a proton of mass $$m_{1}$$ and kinetic energy $$k$$ is given by

$$\lambda_{1} = \dfrac {h}{\sqrt {2m_{1}k}}$$ $$(\because p = \sqrt {2mk})$$

$$\lambda_1= \dfrac {h}{\sqrt {2m_{1}qV}} .... (i)$$ $$ [\because k = qV]$$

For an alpha particle mass $$m_{2}$$ carrying charge $$q_{0}$$ is accelerated through potential $$V$$, then

$$\lambda_{2} = \dfrac {h}{\sqrt {2m_{2}q_{0}V}}$$

$$\because$$ For $$\alpha - particle\ \ (^{4}_{2}He)$$ : $$ q_{0} = 2q$$ and $$m_{2} = 4m_{1}$$

$$\therefore \lambda_{2} = \dfrac {h}{\sqrt {2\times 4m_{1}\times 2q\times V}} .... (ii)$$

The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get
$$\dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {h}{\sqrt {2m_{1}qV}} \times \dfrac {\sqrt {2\times m_{1} \times 4\times 2qV}}{h} = \dfrac {4}{\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}}$$

We get $$\dfrac{\lambda_1}{\lambda_2}= 2\sqrt {2}$$
Question 4
1 / -0

The de Broglie wavelength of an electron is $$0.4\times { 10 }^{ -10 }m$$ when its kinetic energy is $$1.0keV$$. Its wavelength will be $$1.0\times { 10 }^{ -10 }m$$, when its kinetic energy is

A
$$0.2keV$$

B
$$0.8keV$$

C
$$0.63keV$$

D
$$0.16keV$$

Solution

$$\lambda =\cfrac { h }{ p } ,\lambda =\cfrac { h }{ \sqrt { 2mk } } $$
so, $$\lambda \propto \cfrac { 1 }{ \sqrt { k } } $$
$$\Rightarrow \cfrac { 0.4\times { 10 }^{ -10 } }{ 1.0\times { 10 }^{ -10 } } =\cfrac { \sqrt { k } }{ \sqrt { 1 } } \Rightarrow k=0.16keV$$
Question 5
1 / -0

A particle is dropped from a height '$$H$$'. The de Broglie wavelength of the particle depends on height as

A
$$H$$

B
$${ H }^{ -{ 1 }/{ 2 } }$$

C
$${ H }^{ 0 }$$

D
$${ H }^{ { 1 }/{ 2 } }$$

Solution

Let the velocity of particle just before it reaches the ground be $$V$$.
Using energy conservation of the particle, we get
$$\dfrac{1}{2}mV^2 = mgH$$
$$\implies$$ $$V = \sqrt{2gH}$$
de Broglie wavelength of the particle $$\lambda = \dfrac{h}{mV}$$
$$\therefore$$ $$\lambda = \dfrac{h}{m \sqrt{2gH}}$$
$$\implies$$ $$\lambda \propto H^{-1/2}$$
Question 6
1 / -0

A photon will have less energy, if its :

A
Wavelength is longer

B
Wavelength is shorter

C
Amplitude is less

D
Data insufficient

Solution

$$E = \dfrac{hc}{\lambda} \Rightarrow E \propto \dfrac{1}{\lambda}$$
Hence, $$E$$ is less if $$\lambda$$ is longer.
Question 7
1 / -0

An isotropic point source emits light with wavelength $$500 nm$$. The radiation power of the source is $$P=10 W$$. Find the number of photons passing through unit area per second at a distance of $$3 m$$ from the source.

A
$$5.92\times { { 10 }^{ 17 } }/{ { m }^{ 2 }s }$$

B
$$2.23\times { { 10 }^{ 17 } }/{ { m }^{ 2 }s }$$

C
$$2.23\times { { 10 }^{ 18 } }/{ { m }^{ 2 }s }$$

D
$$5.92\times { { 10 }^{ 18 } }/{ { m }^{ 2 }s }$$

Solution

Given, $$\lambda =500nm=500\times { 10 }^{ -9 }m$$
As, $$P={ n }_{ 0 }\dfrac { hc }{ \lambda } $$
$${ n }_{ 0 }=\dfrac { P\lambda }{ hc }$$ .....(i)
where, $$ { n }_{ 0 }$$ is number of photons per second
At distance $$r$$ from point source, number of photons / area / time
$${ n }^{ \prime }=\dfrac { { n }_{ 0 } }{ 4\pi { r }^{ 2 } } =\dfrac { P\lambda }{ hc\cdot 4\pi { r }^{ 2 } } $$ [from equation (i)]
$$=\dfrac { 10\times 500\times { 10 }^{ -9 } }{ 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 }\times 4\pi { \left( 3 \right) }^{ 2 } }$$
$$ =2.23\times { { 10 }^{ 17 } }/{ { m }^{ 2 }s }$$
Question 8
1 / -0

A photon and an electron possesses same de-Broglie wavelength. Given that C$$=$$ speed of light and v$$=$$ speed of electron, which of the following relation is correct? (Here, $$E_e=K.E$$ of electron, $$E_{Ph}=K.E$$ of photon, $$P_e=$$ momentum of electron, $$P_{ph}=$$ momentum of photon).

A
$$\displaystyle\frac{P_e}{P_{Ph}}=\frac{C}{2v}$$

B
$$\displaystyle\frac{E_e}{E_{Ph}}=\frac{C}{2v}$$

C
$$\displaystyle\frac{E_{ph}}{E_e}=\frac{2c}{v}$$

D
$$\displaystyle\frac{P_e}{P_{Ph}}=\frac{2c}{v}$$

Solution

We have, $$\displaystyle \lambda_{Ph}=\frac{h}{P_{Ph}}$$ and $$\displaystyle \lambda_e=\frac{h}{P_e}$$
Given, $$\displaystyle\lambda_{Ph}=\lambda_e$$
$$\therefore$$ We get, $$P_{Ph}=P_e$$
$$\displaystyle\frac{h}{\lambda_{Ph}}=mv$$
$$\therefore \displaystyle\frac{hc}{\lambda_{Ph}}=mcv=\frac{1}{2}mv^2\left(\displaystyle\frac{2c}{v}\right)$$
or $$\displaystyle\frac{E_{Ph}}{E_e}=\frac{2c}{v}$$.
Question 9
1 / -0

The postulate on which the photoelectric equation is derived is :

A
Light is emitted only when electrons jump between orbits

B
Light is absorbed in quanta of energy $$E = hv$$

C
Electrons are restricted to orbits of angular momentum $$n\dfrac {h}{2\pi}$$ where $$n$$ is an integer

D
Electrons are associated with wave of wavelength $$\lambda = \dfrac {h}{p}$$ where $$p$$ is momentum

Solution

When light of sufficiently low wavelength fall on a metal surface, electrons are ejected. This phenomenon is called the photo electric effect and can be understood in terms of the particle nature of light.
i.e. If $$v$$ is the frequency of light that falls on metallic surface, which eject an electron, then we can say that $$E = hv$$ quanta of energy is absorbed during photoelectric emission.
Question 10
1 / -0

The de-Broglie wavelength '$$\lambda$$' of a particle

A
Is proportional to mass

B
Is proportional to impulse

C
Is inversely proportional to impulse

D
Does not depend on impulse

Solution

De Broglie wavelength of the particle $$\lambda = \dfrac{h}{mv}$$
We define impulse as the change in the momentum of the particle i.e. $$I = mv$$
We get $$\lambda = \dfrac{h}{I}$$
$$\implies$$ $$\lambda \propto \dfrac{1}{I}$$

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Dual Nature of Radiation and Matter Test - 38

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Dual Nature of Radiation and Matter Test - 38

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Question 1

$$ 6.4 \times 10^{-34} Js, 2.0$$ eV

$$ 6.0 \times 10^{-34} Js, 2.0$$ eV

$$ 6.4 \times 10^{-34} Js, 3.2 $$ eV

$$ 6.0 \times 10^{-34} Js, 3.2 $$ eV

Solution

Question 2

$$2 : 1$$

$$2\sqrt {2} : 1$$

$$4 : 1$$

$$1 : 2$$

Solution

Question 3

$$2\sqrt {2}$$

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

$$2$$

$$\sqrt {2}$$

Solution

Question 4

$$0.2keV$$

$$0.8keV$$

$$0.63keV$$

$$0.16keV$$

Solution

Question 5

$$H$$

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

$${ H }^{ 0 }$$

$${ H }^{ { 1 }/{ 2 } }$$

Solution

Question 6

Wavelength is longer

Wavelength is shorter

Amplitude is less

Data insufficient

Solution

Question 7

$$5.92\times { { 10 }^{ 17 } }/{ { m }^{ 2 }s }$$

$$2.23\times { { 10 }^{ 17 } }/{ { m }^{ 2 }s }$$

$$2.23\times { { 10 }^{ 18 } }/{ { m }^{ 2 }s }$$

$$5.92\times { { 10 }^{ 18 } }/{ { m }^{ 2 }s }$$

Solution

Question 8

$$\displaystyle\frac{P_e}{P_{Ph}}=\frac{C}{2v}$$

$$\displaystyle\frac{E_e}{E_{Ph}}=\frac{C}{2v}$$

$$\displaystyle\frac{E_{ph}}{E_e}=\frac{2c}{v}$$

$$\displaystyle\frac{P_e}{P_{Ph}}=\frac{2c}{v}$$

Solution

Question 9

Light is emitted only when electrons jump between orbits

Light is absorbed in quanta of energy $$E = hv$$

Electrons are restricted to orbits of angular momentum $$n\dfrac {h}{2\pi}$$ where $$n$$ is an integer

Electrons are associated with wave of wavelength $$\lambda = \dfrac {h}{p}$$ where $$p$$ is momentum

Solution

Question 10

Is proportional to mass

Is proportional to impulse

Is inversely proportional to impulse

Does not depend on impulse

Solution

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