Question

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

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

Answer 1

Answer: (A)

Step I : Determine binding energy of electron :
As we know $ E = \frac{- 1 3 \text{.} 6 \left(\text{Z}\right)^{2}}{n^{2}} = \frac{- 1 3 \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.
∴ $T_{ e } = \frac{ h ⁡ c}{\lambda } - E ⁡_{b}$
$= \frac{1 2 4 2 n \text{meV}}{1 8 n m } - 5 4 \text{.} 4 \text{eV}$
= (69 - 54.4) eV = 14.6 eV
= 23.36 × 10^-19 joule
∴ $ T _{ e ⁡} = \frac{1}{2} \text{mv}^{2}$
$ v = \sqrt{\frac{2 T e ⁡}{m ⁡}}$
$= \sqrt{\frac{2 \times 2 3 \text{.} 3 6 \times 1 0^{- 1 9}}{9 \text{.} 1 \times 1 0^{- 3 1}}} \text{m/s} = 2 \text{.} 3 \times 1 0^{6} \text{m/s}$