Question

If series limit (in nm) of a series of hydrogen spectrum which contains a photon of wavelength 1278.35 nm is expressed as $\frac{1240}{x},$ then $x$ is:

(Given : $hc$ $=$ 1240 eV-nm)

• A. $1.51eV$
• B. $2eV$
• C. $3.4eV$
• D. $0.66eV$

Answer 1

The correct option is A $1.51eV$

If series limit (in nm) of a series of hydrogen,
$E=\frac{hc}{\lambda }\Rightarrow E=\frac{12400eV\dot{A}}{12783.5\dot{A}}=0.97eV$
Hence, the photon will belong to the Paschen series $(0.66\le E\le 1.51)$
Hence, series limit:
$\Rightarrow n=\infty ton=3\therefore \lambda =\frac{hc}{E}=\frac{12400}{1.51}$