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