Question

In hydrogen atom when an electron jumps from second to first orbit, the wavelength of line emitted is :

• A. $0.563, ??
• B. $4861, ??
• C. $4102\ ??
• D. $1213, ??

Answer 1

Answer: (D)

$1213, ??

Answer 2

For ionisation energy of electron in $n$th orbit
$E_{n}=\frac{-13.6}{n^{2}} eV$
For an electron in Ist orbit, i.e., $n=1$
$E_{1}=\frac{-13.6}{1^{2}}=-13.6\, eV$
For an electron in IInd orbit, i.e, $n=2$
$E_{2}=\frac{-13.6}{2^{2}}=-3.4\, eV$
Therefore, energy released
$\Delta E=E_{2}-E_{1}$
$=(-3.4)-(-13.6)$
$=10.2\, eV$
Therefore, wavelength of line emitted is given by
$\lambda =\frac{12375}{\Delta E} ??$=\frac{12375}{10.2} ??
$=1213, ??