Question

In hydrogen atom, an electron jumps from third excited state to ground state, then frequency of emitted light will be
[Take, $h=6.63\times 10^{-34}J-s$ ]

• A. $1.5\times 10^{14}Hz$
• B. $3\times 10^{15}Hz$
• C. $4\times 10^{13}Hz$
• D. $6.5\times 10^{12}Hz$

Answer 1

Answer: (B)

Energy of electron in $n$th orbit in $H$-atom,
$E_n=\frac{-13.6}{n^2}eV$
For third excited state, $n=4$
$\therefore E_4=\frac{-13.6}{4^2}=-0.85eV$
For ground state, $n=1$
$\therefore E_1=\frac{-13.6^2}{1^2}=-13.6eV$
If $v$ be the frequency of emitted photon, then
$hv=E_4-E_1$
$=-0.85-(-13.6)$
$=-0.85+13.6=12.75eV$
$v=\frac{12.75\times 1.6\times 10^{-19}}{h}$
$=\frac{12.75\times 16\times 10^{-19}}{6.63\times 10^{-34}}=3\times 10^{15}Hz$