Discussion Forum



1)

In the hydrogen emission spectrum, for any series, the principal quantum number is n. Corresponding maximum wavelength  $\lambda$  is (R= Rydberg's constant )


A) $\frac{R(2n+1)}{n^{2}(n+1)}$

B) $\frac{n^{2}(n+1)^{2}}{R(2n+1)}$

C) $\frac{n^{2}(n+1)^{}}{R(2n+1)}$

D) $\frac{R(2n+1)}{n^{2}(n+1)^{2}}$

Answer:

Option B

Explanation:

 In hydrogen emission spectrum , wavelength is given by

$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For maximum wavelength of any principal quantum number n,

n1 =n and n2= n+1

$\therefore$  $\frac{1}{\lambda_{max}}=R\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)$

 $=R\left[\frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}}\right]$

$=R\frac{(n^{2}+2n+1-n^{2})}{n^{2}(n+1)^{2}}$

 $\frac{1}{\lambda_{max}}=\frac{R^{2}(2n+1)}{n^{2}(n+1)^{2}}$

$\lambda_{max}=\frac{n^{2}(n+1)^{2}}{R^{}(2n+1)}$