Q

# Help me understand! The energy absorbed in the transition of the electron from n=2 to 1 in H-atom will be

The energy absorbed in the transition of the electron from n=2 to 1 in H-atom will be

• Option 1)

10.2 meV

• Option 2)

3.4 meV

• Option 3)

13.6 meV

• Option 4)

12.75 meV

91 Views

As we learnt in

Lyman Series spectrum -

$\frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )$

Where $n_{1}=1\ and\ n_{2}=2, 3, 4.....$

This lies in Ultraviolet region

-

$\frac{1}{X}=\:RZ^{2}(\frac{1}{n^{2}}-\frac{1}{n{_{2}}^{2}})$

where $n_{1}=1$ for Lyman series

$n_{2}=2 (given)$

Z=1 (hydrogen)

Rydberg constant, R = $1.097\times10^{7}\:m^{-1}$

$\frac{1}{\lambda }\:=\:1.097\times10^{7}\times 1 \times (1- \frac{1}{4})$

$\frac{1}{\lambda }\:=\: \frac{1}{1.212 \times 10^{-7}\:m}$

Energy absorbed = $\frac{hc}{\lambda }$

=        $6.626 \times\: \frac{10^{-34}\times 3 \times 10^{8}}{1.212 \times 10^{-7} \times 1.6 \times 10^{-16}}\:eV$= 10.2 meV

Option 1)

10.2 meV

Option 2)

3.4 meV

Option 3)

13.6 meV

Option 4)

12.75 meV

Exams
Articles
Questions