Physics Q42 – JEE-MAIN 2026

Home Physics Dual Nature of MatterQuestion #69

For a certain metal, when monochromatic light of wavelength

λ

is incident, the stopping potential for photoelectrons is

3 V_{o}

. When the same metal is illuminated by light of wavelength

2 λ

, then the stopping potential becomes

V_{o}

. The threshold wavelength for photoelectric emission for the given metal is

α λ

. The value of

α

is _______.

Clue 1 of 3: Photoelectric Effect Principle

The kinetic energy of emitted photoelectrons depends on the incident photon energy and the work function of the metal.

🧠 Full Solution Path

Step 1: Formulate Einstein's Photoelectric Equation for both cases✦ Active

According to Einstein's photoelectric equation, the stopping potential $V_{s}$ is related to the incident wavelength $λ$ and threshold wavelength $λ_{0}$ by $e V_{s} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$ . We apply this to the two given scenarios:

e (3 V_{o}) = \frac{h c}{λ} - \frac{h c}{λ_{0}} (1)

e V_{o} = \frac{h c}{2 λ} - \frac{h c}{λ_{0}} (2)

Step 2: Solve the system of equations○ Expand

Subtract Equation (2) from Equation (1) to eliminate the work function term ( $\frac{h c}{λ_{0}}$ ):

3 e V_{o} - e V_{o} = (\frac{h c}{λ} - \frac{h c}{λ_{0}}) - (\frac{h c}{2 λ} - \frac{h c}{λ_{0}})

2 e V_{o} = \frac{h c}{λ} - \frac{h c}{2 λ}

2 e V_{o} = \frac{h c}{2 λ}

4 e V_{o} = \frac{h c}{λ} (3)

Now substitute Equation (3) into Equation (2):

e V_{o} = \frac{1}{2} (\frac{h c}{λ}) - \frac{h c}{λ_{0}}

e V_{o} = \frac{1}{2} (4 e V_{o}) - \frac{h c}{λ_{0}}

e V_{o} = 2 e V_{o} - \frac{h c}{λ_{0}}

\frac{h c}{λ_{0}} = e V_{o} (4)

💡 Teacher's Secret Hint

Carefully handle the algebraic manipulation to avoid errors in signs or fractions.

Step 3: Determine the threshold wavelength and the value of

α

○ Expand

We have expressions for $\frac{h c}{λ}$ and $\frac{h c}{λ_{0}}$ from Equations (3) and (4). Divide Equation (3) by Equation (4):

\frac{h c / λ}{h c / λ_{0}} = \frac{4 e V_{o}}{e V_{o}}

\frac{λ_{0}}{λ} = 4

λ_{0} = 4 λ

Given that the threshold wavelength is $α λ$ , we can equate the two expressions:

α λ = 4 λ

α = 4

💡 Teacher's Secret Hint

Remember that the work function and threshold wavelength are properties of the metal and remain constant.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 42 – JEE-MAIN 2026

Clue 1 of 3: Photoelectric Effect Principle

🧠 Full Solution Path