Question

$13.6eV$ is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs $1.50$ times as much energy as the minimum energy required for it to escape from the atom. What is the wavelength and velocity of the emitted electron?

• A. $v=4.0\times {10}^{6}m/s;\lambda =4.7\times {10}^{-10}m$
• B. $v=1.55\times {10}^{6}m/s;\lambda =4.7\times {10}^{-10}m$
• C. $v=1.55\times {10}^{6}m/s;\lambda =16.0\times {10}^{-12}m$
• D. $v=4.0\times {10}^{6}m/s;\lambda =16.0\times {10}^{-12}m$

Answer 1

The correct option is B $v=1.55\times {10}^{6}m/s;\lambda =4.7\times {10}^{-10}m$

1.5 times of 13.6 eV, i.e., 20.4 eV, is absorbed by the hydrogen atom out of which 6.8 eV (20.4 - 13.6) is convereted to kinetic energy. $KE=6.8eV=6.8\times (1.602\times {10}^{-19}coulomb)\times \left(1volt\right)=1.09\times {10}^{-18}J$
Now, $KE=\frac{1}{2}m{v}^2$
or $v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.09\times {10}^{-18}J}{(9.1\times {10}^{-31}kg)}}=1.55\times {10}^{6}m/s.$
From De Broglie equation we have :
$\lambda =\frac{h}{mv}=\frac{(6.63\times {10}^{-34}J.s)}{(9.1\times {10}^{-31}kg)(1.55\times {10}^{6}m/s)}$
$=4.70\times {10}^{-10}m$