At an incident radiation frequency of v1, which is greater than the threshold frequency, the stopping potential for a certain metal is V1. At frequency 2v1, the stopping potential is 3V1. If the stopping potential at frequency 4v1 is n V1, then n is ………

According to Einstein's equation, hv=hv0+eV⇒ V=hev−v0Given V1=hev1−v0 and 3V1=he2v1−v0⇒ 3V1V1=2v1−v0v1−v0⇒v1=2v0∴ V1=he2v0−v0=hev0At frequency 4v1, nV1=he4v1−v0 =he8v0−v0=7×hev0⇒ nV1=7V1⇒n=7

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

At an incident radiation frequency of v₁, which is greater than the threshold frequency, the stopping potential for a certain metal is V₁. At frequency 2v₁, the stopping potential is 3V₁. If the stopping potential at frequency 4v₁ is n V₁, then n is ………

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 7.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

According to Einstein's equation, $hv = {hv}_{0} + eV$

$\Rightarrow V = \frac{h}{e} (v - v_{0})$

Given $V_{1} = \frac{h}{e} (v_{1} - v_{0}) and 3 V_{1} = \frac{h}{e} (2 v_{1} - v_{0})$

$\begin{array}{l} \Rightarrow \frac{3 V_{1}}{V_{1}} = \frac{2 v_{1} - v_{0}}{v_{1} - v_{0}} \Rightarrow v_{1} = 2 v_{0} \\ ∴ V_{1} = \frac{h}{e} (2 v_{0} - v_{0}) = \frac{h}{e} v_{0} \end{array}$