At an incident radiation frequency of v_1,whi | vedclass

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(D) According to Einstein's photoelectric equation,$h v = \phi + K.E_{max}$,where $K.E_{max} = e V_s$.For frequency $v_1$: $h v_1 = \phi + e V_1$  ---

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$7$

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(D) According to Einstein's photoelectric equation,$h v = \phi + K.E_{max}$,where $K.E_{max} = e V_s$.For frequency $v_1$: $h v_1 = \phi + e V_1$  --- $(1)$For frequency $2 v_1$: $h(2 v_1) = \phi + e(3 V_1) = \phi + 3 e V_1$ --- $(2)$From $(1)$,$h v_1 = \phi + e V_1$. Substituting this into $(2)$:$2(\phi + e V_1) = \phi + 3 e V_1$$2 \phi + 2 e V_1 = \phi + 3 e V_1$$\phi = e V_1$Now,substitute $\phi = e V_1$ back into $(1)$:$h v_1 = e V_1 + e V_1 = 2 e V_1$For frequency $4 v_1$,let the stopping potential be $V_s = n V_1$:$h(4 v_1) = \phi + e(n V_1)$$4(h v_1) = e V_1 + n e V_1$$4(2 e V_1) = e V_1(1 + n)$$8 e V_1 = e V_1(1 + n)$$8 = 1 + n$$n = 7$

At an incident radiation frequency of $v_1$,which is greater than the threshold frequency,the stopping potential for a certain metal is $V_1$. At frequency $2 v_1$,the stopping potential is $3 V_1$. If the stopping potential at frequency $4 v_1$ is $n V_1$,then $n$ is

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