JEE Main 2026 — Physics Modern Physics

2026mcqmedium

For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3V_0$ . When the same metal is illuminated by light of wavelength $2\lambda$ , then the stopping potential becomes $V_0$ . The threshold wavelength for photoelectric emission for the given metal is $\alpha\lambda$ . The value of $\alpha$ is _______.

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$1$
B
$4$
C
$2$
D
$3$

Did you get this right?

Solution

Using Einstein's photoelectric equation:

$eV_s = \dfrac{hc}{\lambda} - \Phi$

For the first case:

$3eV_0 = \dfrac{hc}{\lambda} - \Phi$

For the second case:

$eV_0 = \dfrac{hc}{2\lambda} - \Phi$

Multiplying the second equation by $3$ and equating with the first equation:

$\dfrac{hc}{\lambda} - \Phi = \dfrac{3hc}{2\lambda} - 3\Phi$

$2\Phi = \dfrac{3hc}{2\lambda} - \dfrac{hc}{\lambda}$

$2\Phi = \dfrac{hc}{2\lambda}$

$\Phi = \dfrac{hc}{4\lambda}$

Since the work function $\Phi = \dfrac{hc}{\lambda_0}$ , where $\lambda_0$ is the threshold wavelength:

$\dfrac{hc}{\lambda_0} = \dfrac{hc}{4\lambda}$

$\lambda_0 = 4\lambda$

Given $\lambda_0 = \alpha\lambda$ , we get $\alpha = 4$ .

Answer: $4$

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JEE Main 2026 — Physics Modern Physics

2026mcqmedium

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$1$
B
$4$
C
$2$
D
$3$

Did you get this right?

Solution

Using Einstein's photoelectric equation:

$eV_s = \dfrac{hc}{\lambda} - \Phi$

For the first case:

$3eV_0 = \dfrac{hc}{\lambda} - \Phi$

For the second case:

$eV_0 = \dfrac{hc}{2\lambda} - \Phi$

Multiplying the second equation by $3$ and equating with the first equation:

$\dfrac{hc}{\lambda} - \Phi = \dfrac{3hc}{2\lambda} - 3\Phi$

$2\Phi = \dfrac{3hc}{2\lambda} - \dfrac{hc}{\lambda}$

$2\Phi = \dfrac{hc}{2\lambda}$

$\Phi = \dfrac{hc}{4\lambda}$

Since the work function $\Phi = \dfrac{hc}{\lambda_0}$ , where $\lambda_0$ is the threshold wavelength:

$\dfrac{hc}{\lambda_0} = \dfrac{hc}{4\lambda}$

$\lambda_0 = 4\lambda$

Given $\lambda_0 = \alpha\lambda$ , we get $\alpha = 4$ .

Answer: $4$

Your note

JEE Main Modern Physics in other years

Keep practicing Modern Physics

Work through every JEE Main Modern Physics PYQ, year by year.

Browse all Physics topics

JEE Main 2026 — Physics Modern Physics

Options

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Solution

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JEE Main Modern Physics in other years

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JEE Main 2026 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics