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Dual Nature of Radiation question

2016 · Shift 0 · Q40

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Physics
Dual Nature of Radiation
2016 · Shift 0 · Q40

Dual Nature of Radiation question

2016 · Shift 0 · Q40

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

Radiation of wavelength

\lambda ,

is incident on a photocell. The fastest emitted electron has speed

v.

If the wavelength is changed to

{{3\lambda } \over 4},

the speed of the fastest emitted electron will be:

A
$= v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$
B
$= v{\left( {{3 \over 4}} \right)^{{1 \over 2}}}$
C
$\gt v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$
D
$\lt v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}$

View written solutionFree

Correct answer: C

Use Einstein’s photoelectric equation

For the fastest emitted electron,

K_{\max}=\frac{1}{2}mv^2=\frac{hc}{\lambda}-\phi

where $\phi$ is the work function of the metal.

So initially,

\frac{1}{2}mv^2=\frac{hc}{\lambda}-\phi \qquad ...(1)

When wavelength changes to $\dfrac{3\lambda}{4}$

The new photon energy is

\frac{hc}{3\lambda/4}=\frac{4hc}{3\lambda}

Hence the new maximum kinetic energy is

\frac{1}{2}mv'^2=\frac{4hc}{3\lambda}-\phi \qquad ...(2)

Compare with $v\sqrt{\frac{4}{3}}$

From (1),

\frac{1}{2}mv^2=\frac{hc}{\lambda}-\phi

Multiplying by $\frac{4}{3}$ ,

\frac{1}{2}m\left(v\sqrt{\frac{4}{3}}\right)^2=\frac{4}{3}\left(\frac{hc}{\lambda}-\phi\right) =\frac{4hc}{3\lambda}-\frac{4\phi}{3}

But actually from (2),

\frac{1}{2}mv'^2=\frac{4hc}{3\lambda}-\phi

Now compare:

\frac{4hc}{3\lambda}-\phi quad \text{vs} \quad \frac{4hc}{3\lambda}-\frac{4\phi}{3}

Since $\phi>0$ ,

-\phi > -\frac{4\phi}{3}

Therefore,

\frac{1}{2}mv'^2 > \frac{1}{2}m\left(v\sqrt{\frac{4}{3}}\right)^2

which gives

v' > v\sqrt{\frac{4}{3}}

Check options

A: $v' = v\sqrt{\frac{4}{3}}$ ❌ not exact
B: $v' = v\sqrt{\frac{3}{4}}$ ❌ wrong
C: $v' > v\sqrt{\frac{4}{3}}$ ✅ correct
D: $v' < v\sqrt{\frac{4}{3}}$ ❌ wrong

Hence the correct option is C.

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