Question

In Balmer series for hydrogen atom, find the energy of photon corresponding to longest wavelength.

• A. $18.9eV$
• B. $3.03eV$
• C. $1.89eV$
• D. $30.3eV$

Answer 1

Answer: (C)

$1.89eV$

Photon of longest wavelength in Balmer series when a transition takes place from $n_2=3$ to $n_1=2$.
Thus energy of photon emitted $E=-13.6(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ eV
$\therefore$ $E=-13.6(\frac{1}{3^2}-\frac{1}{2^2})$ eV
$⟹E=1.89$ eV