A photoemissive substance is illuminated with a radiation of wavelength \lambda_i so that it releases electrons with de-Broglie wavelength \lambda_e . The longest wavelength of radiation that can emit photoelectron is \lambda_o . Expression for de-Broglie wavelength is given by: (m: mass of the electron, h: Planck's constant and c: speed of light)

\lambda_e=\sqrt\frach2 mc(\frac1\lambda_i-\frac1\lambda_o)

\lambda_e = \frac\sqrth \lambda_i\sqrt2mc

\lambda_e = \frach\sqrt2mc ( \frac1\lambda_i - \frac1\lambda_o )

\lambda_e=\sqrt\frach \lambda_02 m c

Dual Nature of Radiation PYQ — JEE Main 2025 · 7 Apr · Shift 2 · Q52 with Solution

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Correct answer: C

Use Einstein’s photoelectric equation

For incident radiation of wavelength $\lambda_i$ ,

\frac{hc}{\lambda_i}=\phi+K_{\,\max}

where $\phi$ is the work function.

For the threshold (longest) wavelength $\lambda_o$ ,

\phi=\frac{hc}{\lambda_o}

So,

K_{\max}=\frac{hc}{\lambda_i}-\frac{hc}{\lambda_o} =hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)

Relate kinetic energy to momentum

For the emitted electron,

K_{\max}=\frac{p^2}{2m}

Hence,

p=\sqrt{2mK_{\max}} =\sqrt{2m\,hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}

Use de Broglie relation

The de Broglie wavelength is

\lambda_e=\frac{h}{p}

Substituting $p$ :

\lambda_e=\frac{h}{\sqrt{2m\,hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}

Now simplify by taking $h$ inside the square root:

\lambda_e=\sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}

Match with options

This matches:

\boxed{\lambda_e=\sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}}

which is Option C.

Check other options briefly

A: Ignores threshold wavelength $\lambda_o$ , so incorrect.
B: Missing the square root over $h$ in the simplified form; dimensionally incorrect.
C: Correct.
D: Depends only on $\lambda_o$ and not on incident wavelength $\lambda_i$ , so incorrect.

More from Dual Nature of Radiation

View written solutionFree

Correct answer: C

Use Einstein’s photoelectric equation

For incident radiation of wavelength $\lambda_i$ ,

\frac{hc}{\lambda_i}=\phi+K_{\,\max}

where $\phi$ is the work function.

For the threshold (longest) wavelength $\lambda_o$ ,

\phi=\frac{hc}{\lambda_o}

So,

K_{\max}=\frac{hc}{\lambda_i}-\frac{hc}{\lambda_o} =hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)

Relate kinetic energy to momentum

For the emitted electron,

K_{\max}=\frac{p^2}{2m}

Hence,

p=\sqrt{2mK_{\max}} =\sqrt{2m\,hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}

Use de Broglie relation

The de Broglie wavelength is

\lambda_e=\frac{h}{p}

Substituting $p$ :

\lambda_e=\frac{h}{\sqrt{2m\,hc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}

Now simplify by taking $h$ inside the square root:

\lambda_e=\sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}

Match with options

This matches:

\boxed{\lambda_e=\sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}}

which is Option C.

Check other options briefly

A: Ignores threshold wavelength $\lambda_o$ , so incorrect.
B: Missing the square root over $h$ in the simplified form; dimensionally incorrect.
C: Correct.
D: Depends only on $\lambda_o$ and not on incident wavelength $\lambda_i$ , so incorrect.

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