Question

Photon having energy equivalent to the binding energy of 4th state of $H{e}^{+}$ ion is used to eject an electron from the metal with K.E. 2eV. If electron is further accelerated through the potential difference of 4V then the minimum value of de-Broglie wavelength (in $\stackrel{\text{o}}{\text{A}}$) associated with the electron is:
$\left(h=6.6\times {10}^{-34}J-s,m_e=9.1\times {10}^{-31}kg,1ev=1.6\times {10}^{-19}J\right)$

Answer 1

Binding energy of $4^{th}$ state of ${He}^{+}$ ion

we know Rydberg's equation here ${He}^{+}$ ion $\to z=2$

binding energy $\to n_2=\propto$

given $\to n_1=4$

$\frac{1}{\lambda }=R\left(z^2\right)$ $(\frac{1}{n_1^2}-\frac{}{n_2^2})$

$\frac{1}{\lambda }=R\left(2\right)(\frac{1}{4^2}-\frac{1}{{\propto }^2})$

$\frac{1}{\lambda }=\frac{R}{4}$

$\lambda =\frac{4}{R}$

$\Rightarrow energyofphoton=\frac{hc}{\lambda }$

$=hc\left(\frac{R}{4}\right)$

$=5.43\times {10}^{-19}Joules$

$\lambda =\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{{2mV}_q}}$

$E=2eV$

$V_q=4ev$

${\lambda }_{min}=\frac{6.626\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}}\times 4ev}$

$=\frac{h}{1.079\times {10}^{-24}}=6.139\stackrel{0}{A}$