Question

What is the wavelength of the most energetic photon emitted in the Balmer's series of the hydrogen atom?

• A. $3694A$
• B. $5756A$
• C. $3333A$
• D. $3669A$

Answer 1

Answer: (D)

$3669A$

Balmer series means wavelength is given by $\frac{1}{\lambda }=R(\frac{1}{4}-\frac{1}{n^2})$

for maximum wavelength the $n$ should be max so putting $n\to \infty$

we get $\lambda =\frac{4}{R}$

putting the Rydberg's constant as $R=1.09\times {10}^7/m$

we get $\lambda =3669\times {10}^{-10}meter=3669Angstrom$