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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2010

MCQ (Single Correct Answer)
Statement - $$1$$ : When ultraviolet light is incident on a photocell, its stopping potential is $${V_0}$$ and the maximum kinetic energy of the photoelectrons is $${K_{\max }}$$. When the ultraviolet light is replaced by $$X$$-rays, both $${V_0}$$ and $${K_{\max }}$$ increase.

Statement - $$2$$ : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

A
Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is the correct explanation of Statement - $$1$$
B
Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not the correct explanation of Statement - $$1$$
C
Statement - $$1$$ is is false, Statement - $$2$$ is true
D
Statement - $$1$$ is is true, Statement - $$2$$ is false

Explanation

we know that

$$e{V_0} = {K_{\max }} = hv - \phi $$

where, $$\phi $$ is the work function.

Hence, as v increases (note that frequency of $$X$$-rays is greater than that of $$U.V.$$ rays), both $${V_0}$$ and $${K_{\max }}$$ increase. So statement - $$1$$ is correct
2

AIEEE 2009

MCQ (Single Correct Answer)
The surface of a metal is illuminated with the light of $$400$$ $$nm.$$ The kinetic energy of the ejected photoelectrons was found to be $$1.68$$ $$eV.$$ The work function of the metal is : $$\left( {hc = 1240eV.nm} \right)$$
A
$$1.41$$ $$eV$$
B
$$1.51$$ $$eV$$
C
$$1.68$$ $$eV$$
D
$$3.09$$ $$eV$$

Explanation

$$\lambda = 400nm,\,\,hc = 1240eV.nm,\,\,K.E. = 1.68\,eV$$

We know that $${{hc} \over \lambda } - W = K.E \Rightarrow W = {{hc} \over \lambda } - K.E$$

$$ \Rightarrow W = {{1240} \over {400}} - 1.68$$

$$ = 3.1 - 1.68 = 1.42\,eV$$
3

AIEEE 2008

MCQ (Single Correct Answer)
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

If a strong diffraction peak is observed when electrons are incident at an angle $$'i'$$ from the normal to the crystal planes with distance $$'d'$$ between them (see figure), de Broglie wavelength $${\lambda _{dB}}$$ of electrons can be calculated by the relationship ($$n$$ is an integer)

A
$$d\,\sin \,i = n{\lambda _{dB}}$$
B
$$2d\,\cos \,i = n{\lambda _{dB}}$$
C
$$2d\,\sin \,i = n{\lambda _{dB}}$$
D
$$d\,\cos \,i = n{\lambda _{dB}}$$

Explanation

$$2d\,\cos \,i = n{\lambda _{dB}}$$
4

AIEEE 2008

MCQ (Single Correct Answer)
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).

Electrons accelerated by potential $$V$$ are diffracted from a crystal. If $$d = 1\mathop A\limits^ \circ $$ and $$i = {30^ \circ },\,\,\,V$$ should be about
$$\left( {h = 6.6 \times {{10}^{ - 34}}Js,{m_e} = 9.1 \times {{10}^{ - 31}}kg,\,e = 1.6 \times {{10}^{ - 19}}C} \right)$$

A
$$2000$$ $$V$$
B
$$50$$ $$V$$
C
$$500$$ $$V$$
D
$$1000$$ $$V$$

Explanation

Using Bragg's equation $$2d$$ $$\sin \theta = n\lambda $$

Here $$n=1,$$ $$\theta = 90 - i = 90 - 30 = {60^ \circ }$$

$$\therefore$$ $$2d$$ $$\sin \,\theta \, = \lambda \,\,\,\,\,\,\,\,\,\,....\left( i \right)$$



Also, $$\lambda = {{12.27} \over {\sqrt V }} \times {10^{ - 10}}m\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$(i)$$ & $$(ii)$$ $$2 \times {10^{ - 10}} \times \sin {60^ \circ } = {{12.27} \over {\sqrt V }} \times {10^{ - 10}}$$

$$V = {{{{\left( {12.27} \right)}^2}} \over 3} = 50V$$

Questions Asked from Dual Nature of Radiation

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