Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

What is the de-Broglie wavelength of the $$\alpha$$- particle accelerated through a potential difference $$V$$

A
$$\dfrac{0.287}{\sqrt V}\ A^o$$

B
$$\dfrac{12.27}{\sqrt V}\ A^o$$

C
$$\dfrac{0.101}{\sqrt V}\ A^o$$

D
$$\dfrac{0.202}{\sqrt V}\ A^o$$

Solution

As we know,
$$\lambda =\dfrac {h}{\sqrt{2mE}}=\dfrac {h}{\sqrt{2m_{\alpha}Q_{\alpha}V}}$$
On putting $$Q_{\alpha}=2\times 1.6\times 10^{-19}\ C$$
$$m_{\alpha }=4m_p=4\times 1.67 \times 10^{-27}\ kg\Rightarrow \lambda =\dfrac {0.101}{\sqrt V}\ A^o$$
Question 2
1 / -0

The kinetic energy of electron and proton is $$10^{-32}\ J$$. Then the relation between their de-Broglie wavelength is

A
$$\lambda_p < \lambda_e$$

B
$$\lambda_p > \lambda_e$$

C
$$\lambda_p = \lambda_e$$

D
$$\lambda_p =2 \lambda_e$$

Solution

According to the question,
By using $$\lambda =\dfrac {h}{\sqrt{2mE}}\ E=10^{-32}\ J$$= Constant for both
particles. Hence $$\lambda \propto \dfrac {1}{\sqrt m}$$ Since $$m_p > m_e$$ so $$\lambda_p < \lambda_e$$.
Question 3
1 / -0

If a photon has velocity $$c$$ and frequency $$v$$, then which of the following represents its wavelength

A
$$\dfrac {hc}{E}$$

B
$$\dfrac {hv}{c}$$

C
$$\dfrac {hv}{c^2}$$

D
$$hv$$

Solution

If a photon has velocity c and frequency v, then
$$E=\dfrac{hc}{\lambda}\Rightarrow \lambda =\dfrac{hc}{E}$$
Question 4
1 / -0

The de-Broglie wavelength of a neutron at $$27^oC$$ is $$\lambda$$. What will be its wavelength at $$927^oC$$

A
$$\lambda /2$$

B
$$\lambda /3$$

C
$$\lambda /4$$

D
$$\lambda /9$$

Solution

As we know,
$$\lambda_{neutron } \propto \dfrac {1}{\sqrt T}\Rightarrow \dfrac {\lambda_1}{\lambda_2}=\sqrt{\dfrac {T_2}{T_1}}$$

$$\Rightarrow \dfrac {\lambda}{\lambda_2}=\sqrt{\dfrac{(273+927)}{(273+27)}}=\sqrt{\dfrac{1200}{300}}=2\Rightarrow \lambda_2=\dfrac {\lambda}{2}$$
Question 5
1 / -0

For moving ball of cricket, the correct statement about de-Broglie wavelength is

A
It is not applicable for such big particle

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

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

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

Solution

de Broglie wavelength associated with the particle
$$\lambda =\dfrac hp$$
where h is the Planck's constant
now we know thwt ,
$$E=\dfrac12 mv^2$$ and $$p=mv$$
by mutiplying by m on both sides we get,
$$p=\sqrt{2mE}$$
hence,
$$\lambda =\dfrac h{\sqrt{2mE}}$$
Question 6
1 / -0

The mass of the electron varies with

A
The size of the cathode ray tube

B
The variation of $$'g'$$

C
Velocity

D
Size of the electron

Solution

Mass is basically a constant term for any physical application at low velocity. But in accordance with Einstein's theory of relativity, at higher speeds the mass of the particle change according to formula

$$m=\dfrac{m}{\sqrt{1-(v^2/c^2)}}$$
Question 7
1 / -0

The de-Broglie wavelength is proportional to

A
$$\lambda \propto \dfrac 1v$$

B
$$\lambda \propto \dfrac 1m$$

C
$$\lambda \propto \dfrac 1p$$

D
$$\lambda \propto p$$

Solution

As we know,
$$\lambda =\dfrac hp $$

$$\Rightarrow \lambda \propto \dfrac 1p$$
Question 8
1 / -0

When yellow light is incident on a surface, no electrons are emitted while green light can emit. If red light is incident on the surface then

A
No electron are emitted

B
Photons are emitted

C
Electrons are higher energy are emitted

D
Electrons are lower energy are emitted

Solution

$$\lambda_r > \lambda_y > \lambda_g$$. Here threshold wavelength $$< \lambda_y$$.
hence, no electrons are emitted.
Question 9
1 / -0

de-Broglie wavelength of a body of mass $$m$$ and kinetic energy $$E$$ is given by

A
$$\lambda =\dfrac {h}{mE}$$

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

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

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

Solution

de Broglie wavelength associated with the particle
$$\lambda =\dfrac hp$$
where h is the Planck's constant
now we know thwt ,
$$E=\dfrac12 mv^2$$ and $$p=mv$$
by mutiplying by m on both sides we get,
$$p=\sqrt{2mE}$$
hence,
$$\lambda =\dfrac h{\sqrt{2mE}}$$
Question 10
1 / -0

A proton and an $$\alpha$$ particle are acclerated through a potential difference of $$100\ V$$. The ratio of the wavelength associated with the proton to that associated with an $$\alpha$$- particle is

A
$$\sqrt 2:1$$

B
$$2:1$$

C
$$2\sqrt 2:1$$

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

Solution

As we know,
$$\lambda =\dfrac{h}{\sqrt{2mQV}}\Rightarrow \propto \dfrac{1}{\sqrt{mQ}}\Rightarrow \dfrac{\lambda_p}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha}Q_{\alpha}}{m_p Q_p}}$$
$$=\sqrt{\dfrac{4m_p\times 2Q_p}{m_p\times Q_p}}=2\sqrt 2$$

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Dual Nature of Radiation and Matter Test - 58

Result

Dual Nature of Radiation and Matter Test - 58

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Question 1

$$\dfrac{0.287}{\sqrt V}\ A^o$$

$$\dfrac{12.27}{\sqrt V}\ A^o$$

$$\dfrac{0.101}{\sqrt V}\ A^o$$

$$\dfrac{0.202}{\sqrt V}\ A^o$$

Solution

Question 2

$$\lambda_p < \lambda_e$$

$$\lambda_p > \lambda_e$$

$$\lambda_p = \lambda_e$$

$$\lambda_p =2 \lambda_e$$

Solution

Question 3

$$\dfrac {hc}{E}$$

$$\dfrac {hv}{c}$$

$$\dfrac {hv}{c^2}$$

$$hv$$

Solution

Question 4

$$\lambda /2$$

$$\lambda /3$$

$$\lambda /4$$

$$\lambda /9$$

Solution

Question 5

It is not applicable for such big particle

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

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

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

Solution

Question 6

The size of the cathode ray tube

The variation of $$'g'$$

Velocity

Size of the electron

Solution

Question 7

$$\lambda \propto \dfrac 1v$$

$$\lambda \propto \dfrac 1m$$

$$\lambda \propto \dfrac 1p$$

$$\lambda \propto p$$

Solution

Question 8

No electron are emitted

Photons are emitted

Electrons are higher energy are emitted

Electrons are lower energy are emitted

Solution

Question 9

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

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

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

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

Solution

Question 10

$$\sqrt 2:1$$

$$2:1$$

$$2\sqrt 2:1$$

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

Solution

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