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

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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 ………

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Detailed Solution

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

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

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

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