Question

The wave number of energy emitted when electron jumps from fourth orbit to seconds orbit in hydrohen in $20,497c{m}^{-1}$. The wave number of energy for the same transition in $H{e}^{+}$ is

• A. $5,099c{m}^{-1}$
• B. $20,497c{m}^{-1}$
• C. $40,994c{m}^{-1}$
• D. $81,988c{m}^{-1}$

Answer 1

The correct option is D $81,988c{m}^{-1}$

Given that,

Wave number of energy for hydrogen $=20497c{m}^{-1}$

Now, for hydrogen

$\frac{1}{{\lambda }_H}=R(\frac{1}{2^2}-\frac{1}{4^2})=20497$

Now, for helium

$\frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{4^2})\times 2^2$

$\frac{1}{{\lambda }_{He}}=20497\times 4$

$\frac{1}{{\lambda }_{He}}=81988c{m}^{-1}$

Hence, the wave number of energy for helium is $81988c{m}^{-1}$