Question

Photoelectric emission is observed from a surface for frequencies ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }{y}^{\prime }$ of incident radiation, $(x>y)$. If $x=2y$, then find the threshold frequency in term of ${}^{\prime }{x}^{\prime }$, if ratio of stopping potentials using ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }{y}^{\prime }$ respectively is $4:1$.

Answer 1

$\begin{array}{cc}hx=h{v}_0+K.E_1& x=2y\frac{K.E_1}{K.E_2}=\frac{4}{1}\\ hy=h{v}_0+K.E_2& K.E_1=4K.E_2\\ hx=h{v}_0+4K.E_2\to \left(i\right)& \\ h\left(2x\right)=h{v}_0+K.E_2\to \left(ii\right)& K.E_2=2hx-h{v}_0\\ Solving\left(i\right)\&\left(ii\right)& \\ hx=h{v}_0+4\left(2hx-h{v}_0\right)& \\ hx=h{v}_0+8hx-4h{v}_0& \\ -7hx=-3h{v}_0& \\ x=\frac{3v_0}{7}& \end{array}$