Question

The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is $x × 10^{–8}\ m$. The value of $x$ is ______. (Nearest Integer)
(Given : Energy of the electron in the first shell of the hydrogen atom is $2.2 \times 10^{–18} J$;
$h = 6.63 × 10^{–34}Js$ and $c = 3 × 10^8 ms^{–1}$)

Answer 1

Energy required for ionisation of Li atom,
$= 2.2 × 10^{–18}×\frac 94$
J [Assume this formula is True for Li atom]
$∵ E = \frac {hc}{λ}$

$2.2 \times 10^{-18} \times \frac 94 = \frac {6.63 \times 10-34 \times3 \times 10^8}{λ}$
$λ = 4 \times 10^{-8} m$

So, the answer is $4$.