Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

During an experiment an $$\alpha $$- particle and a proton are having a difference , their de Broglie wavelenght ratio will be

A
$$1:2$$

B
$$1:4$$

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

D
none of these

Solution

Hence, option $$(C)$$ is correct answer.
Question 2
1 / -0

If the de-Broglie wavelengths associated with a proton and $$ \alpha $$ -particle are equal, then the ratio of velocities of the proton and the a-particle will be:

A
$$4:1$$

B
$$2:1$$

C
$$1:2$$

D
$$1:4$$

Solution

$$\begin{array}{l} { \lambda _{ p } }=\dfrac { h }{ { m{ V_{ 1 } } } } \\ { \lambda _{ \alpha } }=\dfrac { h }{ { 4m{ v_{ 2 } } } } \\ \Rightarrow \dfrac { h }{ { m{ v_{ 1 } } } } =\dfrac { h }{ { 4m{ v_{ 2 } } } } \\ \Rightarrow { v_{ 1 } }=4{ v_{ 2 } } \\ \Rightarrow \dfrac { { { v_{ 1 } } } }{ { { v_{ 2 } } } } =4:1 \\ Hence, \\ option\, \, A\, \, is\, correct\, \, answer. \end{array}$$
Question 3
1 / -0

If there are two de-Broglie waves formed in an orbit of H-atom then the energy of electron in that orbit is.

A
$$-3.4$$ $$eV$$

B
$$-13.6$$ $$eV$$

C
$$-9.06$$ $$eV$$

D
$$1.51$$ $$eV$$

Solution

$$\begin{array}{l} 2\pi r=2\cdot \frac { h }{ { mv } } \\ \Rightarrow mvr=\frac { h }{ \pi } =2\left( { \frac { h }{ { 2\pi } } } \right) \\ \Rightarrow n=2 \\ \therefore E=-13.6\times \frac { 1 }{ 4 } \\ =-3.4\, ev \\ Hence,\, option\quad A\quad is\, correct\, answer. \end{array}$$
Question 4
1 / -0

An electron with an initial kinetic energy of $$100\ eV$$ is accelerated through a potential difference of $$50\ V$$. Now the de-Broglie wavelength of electron becomes

A
$$1\ \overset{o}{A}$$

B
$$\sqrt{1.5}\ \overset{o}{A}$$

C
$$\sqrt 3\ \overset{o}{A}$$

D
$$12.27\ \overset{o}{A}$$

Solution

The initial kinetic energy of the electron is $$100eV$$

Now, it is accelerated through a potential is $$50V$$

Therefore, the additional kinetic energy it gains is

$$K'=eV=50\,eV$$

Hence, the total kinetic energy possessed by the electron is $$150\,eV$$ which is equivalent to $$150\times 1.6\times 10^{-19}\,J$$

$$\lambda =\dfrac{h}{\sqrt{2mK}}=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 150\times 1.6\times 10^{-19}}}=1\times 10^{-10}\,m=1\ \overset{o}{A}$$
Question 5
1 / -0

The anode voltage of a photocell is kept fixed. The wavelength of the light falling on the cathode is gradually changed. The plate current $$I$$ of the photocell varies as follows:

A

B

C

D

Solution

As $$\lambda$$ will decrease know will decrease so current should decrease and finally fall to zero when $${ \lambda }_{ 0 }$$ is achieved.
Thus, graph will be this.
Question 6
1 / -0

What should be the momentum (in gram centimetre per second) of a particle if its de-Broglie wavelength is 1 angstrom and the value of his $$6.6252 \times 10 ^ { - 27 }$$ erg second ?

A
$$
6.6252 \times 10 ^ { - 19 } \mathrm { gcm } / \mathrm { s }
$$

B
$$
6.6252 \times 10 ^ { - 21 } \mathrm { gcm } / \mathrm { s }
$$

C
$$
6.6252 \times 10 ^ { - 24 } \mathrm { gcm } / \mathrm { s }
$$

D
$$
6.6252 \times 10 ^ { - 27 } \mathrm { gcm } / \mathrm { s }
$$

Solution

Given that,
$$\lambda = 1 A^o = 1 \times 10^{-8} cm, h= 6.6252 \times 10^{-27} erg\ second$$
or $$ p = \dfrac{6.6252 \times 10^{-27}}{ 10 ^{-8}} = 6.6252 \times 10^{-19} \ gram. cm/sec$$
Question 7
1 / -0

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature $$27^oC$$ and $$127^oC$$ respectively is

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

B
$$2$$

C
$$\sqrt{\dfrac{8}{3}}$$

D
$$1$$

Solution
Question 8
1 / -0

Monochromatic radiation emitted when electron on hydrogen atom jumps from first exited to the grounds state irradiates a photosnesitive material. The stopping potential is measured to be 3.57V. The threshold frequency of the material is :

A
$$1.6\times 10^{15}$$Hz

B
$$2.5\times 10^{15}$$Hz

C
$$4\times 10^{15}$$Hz

D
$$5\times 10^{15}$$Hz

Solution

For hydrogen atom,

$$E_n=-\dfrac{13.6}{n^2}eV$$

For ground state, $$n=1$$

$$E_1=-\dfrac{13.6}{1^2}=-13.6eV$$

For first excited state, $$n=2$$

$$E_2=-\dfrac{13.6}{2^2}=-3.4eV$$

The energy of the emitted photon when an electron jumps from first excited state to ground state is,

$$h\nu=E_2-E_1$$

$$=-3.4eV-(-13.6eV)=10.2eV$$

Maximum kinetic energy,

$$K_{max}=eV_s=e\times 3.57V=3.57eV$$

According to Einstein's photoelectric equation,

$$K_{max}=h\nu-\phi_0$$

where, $$\phi_0$$ is the work function and $$h\nu$$ is the incident energy

$$\phi_0=h\nu-K_{max}$$

$$=10.2eV-3.57eV=6.63eV$$

Threshold frequency,

$$\nu_0=\dfrac{\phi_0}{h}=\dfrac{6.63\times 1.6\times 10^{-19}\,J}{6.63\times 10^{-34}\,Js}=1.6\times 10^{15}\,Hz$$
Question 9
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If a photon having wavelength $$6.2 { nm }$$ was allowed to strike a metal plate having work function $$50 { eV }$$, then calculate wavelength associated with emitted electron.

A
$$1 \times 10 ^ { - 10 } \mathrm { m }$$

B
$$2\times 10 ^ { - 10 } \mathrm { m }$$

C
$$3 \times 10 ^ { - 18 } \mathrm { m }$$

D
$$4 \times 10 ^ { - 14 } \mathrm { m }$$

Solution

Energy of photon $$=\dfrac{12400}{\lambda(in\,A^0)}eV$$

$$=\dfrac{12400}{62}$$

$$=200eV$$

$$KE_{max}=Energy\,\,.of \,\,photon-work function$$

$$=200-50$$

$$=150eV$$

We know,

$$\lambda=\dfrac{h}{\sqrt{2M(KE_{max})}}$$

$$=\dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times (150\times 1.6\times 10^{-19})}}$$

$$=\dfrac{6.6\times 10^{-34}}{66\times 10^{-25}}$$

$$=1\times 10^{-10}m$$

$$=1A^0$$
Question 10
1 / -0

de-Broglie equation is

A
$$n\lambda =2d\quad sin\theta $$

B
$$E=hv$$

C
$$E={ mc }^{ 2 }$$

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

Solution

de-Broglie equation is an expression for wavelength and momentum of a particle.
$$\lambda=\cfrac{h}{mv}$$
Option D is correct.