An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the minimum r

Question

An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the minimum required for it to escape from the atom. What is the speed of the emitted electron?

Accepted Answer

K.E. of emitted electron = 0.5 × 13.6 eV

Now, $K.E. = \frac{1}{2}mv^{2}$

Or, $6.8 \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2}$

$\Rightarrow v = 1.55 \times 10^{6}\text{m/s}$

Question

Options

Solution

More Atomic Structure Questions