Question

The frequency of light emitted, when the electron makes transition from the level of principle quantum number $n=2$ to the level with $n=1$ is (Take, the ionization energy of hydrogen to be $13.6 eV$ and $\left.h \simeq 4 \times 10^{-15} eV - s \right)$

• A. $2.55 \times 10^{15} Hz$
• B. $1.7 \times 10^{15} Hz$
• C. $3.4 \times 10^{15} Hz$
• D. $5.1 \times 10^{15} Hz$

Answer 1

Answer: (A)

As ionisation energy is $13.6 \,eV$, so its negative will be the energy of atom in ground state $(n=1)$
hence $E_{1}=-13.6\, eV$
Energy in second orbit $(n=1)$
$E_{2}=-\frac{13.6}{2^{2}}=-34 \,eV$
When electron jumps from $n=2$ to $n=1$, then energy of photon emitted by atom
$\Delta \,E =E_{2}-E_{1} $
$=-3.4-(-13.6) $
$\Delta\, E =10.2\, eV$
So, frequency (v),
$h v =10.2 \,eV $
$v =\frac{10.2 eV }{h}=\frac{10.2 eV }{4 \times 10^{-15} eV - s } $
$v =255 \times 10^{+15} \,Hz$