Question

Photon having energy equivalent to the binding energy of $4^{th}$ state of $H{e}^{+}$ atom is used to eject an electron from the metal surface of work function 1.4 eV. If electrons are further accelerated through the potential difference of $4V$ then the minimum value of De - broglie wavelength (in $\stackrel{o}{A}$) associated with the electron is: (Given $\sqrt{174.72}=13.24$, take $h=6.62\times {10}^{-34}Js$

Answer 1

Total energy $=\frac{13.6Z^2}{n^2}=\frac{13.6(Z{)}^2}{(4{)}^2}=3.4eV$
$K.E.=3.4-1.4=2eV$
Since electrons are further accelerated through the potential difference of $4V$
Now, Total energy $=2+4=6eV$
For electron, wave length $\left(\lambda \right)=\frac{6.62\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 6\times 1.6\times {10}^{-19}}}$

$\Rightarrow \lambda =5\times {10}^{-10}m$
So, $\lambda =5\stackrel{o}{A}$