1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

When photons of wavelength $${\lambda _1}$$ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength $${\lambda _2}$$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $${\lambda _1}$$ is used then find the stopping potential for this case :
A
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} - {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$
B
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$
C
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {3 \over {2{\lambda _1}}}} \right]$$
D
$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$

Explanation

We know,

Einstein's photoelectric equation,

$$eV = {{hc} \over \lambda } - {\phi _0}$$

and $${\phi _0}$$, $${{hc} \over {{\lambda _0}}}$$, where $${{\lambda _0}}$$ is the threashhold wavelength.

$$ \therefore $$   In first case,

eV $$ = {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}$$      . . .(1)

and in second case,

3 eV $$ = {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$$       . . .(2)

Now, let slopping potential = V1 when light of wavelength $$\lambda $$3 is used then,

eV1 $$ = {{hc} \over {{\lambda _3}}} - {{hc} \over {{\lambda _0}}}$$       . . .(3)

From (1) and (2) get,

$$3\left( {{{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}} \right) = {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$$

$$ \Rightarrow $$   $${{3hc} \over {{\lambda _1}}}$$ $$-$$ $${{hc} \over {{\lambda _2}}}$$ $$=$$ $${{2hc} \over {{\lambda _0}}}$$

$$ \Rightarrow $$   $${{hc} \over {{\lambda _0}}}$$ $$=$$ $${{3hc} \over {2{\lambda _1}}}$$ $$-$$ $${{hc} \over {2{\lambda _2}}}$$

Putting this value of $${{hc} \over {{\lambda _0}}}$$ in equation 3,

eV1 $$ = {{hc} \over {{\lambda _3}}} - {{3hc} \over {2{\lambda _1}}} + {{hc} \over {2{\lambda _2}}}$$

$$ \Rightarrow $$   V1 $$ = {{hc} \over e}$$ $$\left[ {{1 \over {{\lambda _3}}} - {3 \over {2{\lambda _1}}} + {1 \over {2{\lambda _2}}}} \right]$$
2
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

A photoelectric surface is illuminated successively by monochromatic light of wavelengths $$\lambda $$ and $${\lambda \over 2}.$$ If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :
A
$${{hc} \over {3\lambda }}$$
B
$${{hc} \over {2\lambda }}$$
C
$${{hc} \over {\lambda }}$$
D
$${3\,{hc} \over {\lambda }}$$

Explanation

We know,

Einstein's photo electric equation,

(KE)max = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$0

In first case,

K = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$0           . . .(1)

In second case,

3K = $${{2hc} \over \lambda }$$ $$-$$ $$\phi $$0           . . .(2)

$$ \Rightarrow $$   $$3\left( {{{hc} \over \lambda } - {\phi _0}} \right)$$ = $${{2hc} \over \lambda } - {\phi _0}$$

$$ \Rightarrow $$   $${{3hc} \over \lambda }$$ $$-$$ 3$$\phi $$0 = $${{2hc} \over \lambda }$$ $$-$$ $$\phi $$0

$$ \Rightarrow $$   $${{hc} \over \lambda }$$ = 2$$\phi $$0

$$ \Rightarrow $$   $$\phi $$  =  $${{hc} \over {2\lambda }}$$
3
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X–rays. It produces continuous as well as characteristic X-rays. If $$\lambda $$min is the smallest possible wavelength of X-ray in the spectrum, the variation of log$$\lambda $$min with log V is correctly represented in:
A
B
C
D

Explanation

In X-ray tube, $${\lambda _{\min }} = {{hc} \over {eV}}$$

$$\log \left( {{\lambda _{\min }}} \right) = \log \left( {{{hc} \over e}} \right) - \log V$$

Clearly, log ($$\lambda $$min) versus log V graph slope is negative hence option (3) is correct.
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\lambda _A}$$ to $${\lambda _B}$$ after the collision is:
A
$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}$$
B
$${{{\lambda _A}} \over {{\lambda _B}}} = 2$$
C
$${{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}$$
D
$${{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}$$

Explanation

From question, mA = M; mB = $${m \over 2}$$

uA = V and uB = 0

Let after collision velocity of A = V1 and

velocity of B = V2

Applying law of conservation of momentum,

mu = mv1 + $$\left( {{m \over 2}} \right){v_2}$$

$$ \Rightarrow $$ 24= 2v1 + v2 ........(i)

By law of collision

$$e = {{{v_2} - {v_1}} \over {u - 0}}$$

$$ \Rightarrow $$ u = v2 - v1 ..........(ii)

[As collision is elastic, e = 1]

using eqns (i) and (ii)

v1 = $${4 \over 3}$$ and v2 = $${4 \over 3}v$$

We know, de-Broglie wavelength $$\lambda $$ = $${h \over p}$$

$$ \therefore $$ $${{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}}$$ = $${{{m \over 2} \times {4 \over 3}u} \over {m \times {4 \over 3}}}$$ = 2

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