Question

What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum. (take hc = 1240 eV nm)

• A. 102 nm
• B. 150 nm
• C. 82 nm
• D. 122 nm

Answer 1

The correct option is D 122 nm

Find out $\lambda$ of light

$hc=1240eVmm$

Lymen series of the hydrogen spectrum

Solution

For any series the transistion that produces the least energetic photon is the transition between the home base level that defines the series and the level immediately above it. For the hymen series, the home-base level is at $n=1$

So the transition that produces the least energetic photon is the transition from the $n=2$ level to then $n=1$ level.

$\therefore$ The wavelength for the least energetic photon is

$\lambda =\frac{hc}{E_2-E_1}$

Here $hc=1240eVnm$

$E_1=-\frac{13.6}{1^2}eV=-13.6eV$

(as $E_n=-\frac{13.6}{n^2}eV)$

$E_2=-\frac{13.6}{2^2}eV=-3.4eV\therefore \lambda =\frac{1240eVnm}{-3.4eV-(-13.6eV)}=\frac{1240eVnm}{102eV}=122nm$