Question

The velocity of the electrons liberated by electromagnetic radiation of wavelength $\lambda =18.0nm$ , from stationary $H{e}^{+}$ ions in the ground state, is

• A. $2.3\times 10^{6}m{s}^{-1}$
• B. $1.3\times 10^{6}m{s}^{-1}$
• C. $2.3\times 10^{5}m{s}^{-1}$
• D. $1.3\times 10^{3}m{s}^{-1}$

Answer 1

Answer: (A)

Step I : Determine binding energy of electron :
As we know $E=\frac{-13\text{.}6{\left(\text{Z}\right)}^2}{n^2}=\frac{-13\text{.}6{\left(2\right)}^2}{{\left(1\right)}^2}\left(\text{For ground state}\right)$
= - 54.4 eV
∴ The binding energy of electron is E_b = - E = 54.4 eV
Step II : Determine kinetic energy of electron.
∴ $T_e=\frac{hc}{\lambda }-E{}_b$
$=\frac{1242n\text{meV}}{18nm}-54\text{.}4\text{eV}$
= (69 - 54.4) eV = 14.6 eV
= 23.36 × 10^-19 joule
∴ $T_e=\frac{1}{2}{\text{mv}}^2$
$v=\sqrt{\frac{2Te}{m}}$
$=\sqrt{\frac{2\times 23\text{.}36\times 10^{-19}}{9\text{.}1\times 10^{-31}}}\text{m/s}=2\text{.}3\times 10^6\text{m/s}$