Question

In hydrogen atom, an electron in its ground state absorbs two times of the energy as if requires escaping $\left(\right.13.6eV\left.\right)$ from the atom. The wavelength of the emitted electron will be

• A. $1.34\times 10^{- 10}m$
• B. $2.34\times 10^{- 10}m$
• C. $3.34\times 10^{- 10}m$
• D. $4.44\times 10^{- 10}m$

Answer 1

Answer: (C)

Energy absorbed by an atom $=2\times 13.6=27.2eV$

Energy consumed in escape $=13.6eV$

Energy converted into K.E. $=13.6\times 1.6\times 10^{- 19}J$

$\lambda =h / \sqrt{2 m K E}$ $\because $ ( $h=6.6\times 10^{- 34}kgm^{2}s^{- 1}$ and $m=9.1\times 10^{- 31}kg$ )

$=\frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 1 0^{- 31} \times 13.6 \times 1.6 \times 1 0^{- 19}}}$

On solving

$=3.34\times 10^{- 10}m$