Question

In Bohr’s theory of Hydrogen atom, the electron jumps from higher orbit ‘$n$’ to lower orbit ‘$p$’. The wavelength will be minimum for the transition

• A. $n=5$ to $p=4$
• B. $n=4$ to $p=3$
• C. $n=3$ to $p=2$
• D. $n=2$ to $p=1$

Answer 1

Answer: (D)

Wavelength of light emitted due to transition from $n$ to $p$ orbits
$\frac{1}{\lambda }=R(\frac{1}{p^2}-\frac{1}{n^2})$
$\Rightarrow \lambda =\frac{n^2{p}^2}{R\left(n^2-p^2\right)}$
Case $(A):n=5p=4$
We get ${\lambda }_A=\frac{5^2\times 4^2}{R\left(5^2-4^2\right)}=44.44R$
Case $(B):n=4p=3$
We get ${\lambda }_B=\frac{4^2\times 3^2}{R\left(4^2-3^2\right)}=20.57R$
Case $(C):n=3p=2$
We get ${\lambda }_C=\frac{3^2\times 2^2}{R\left(3^2-2^2\right)}=7.2R$
Case $(D):n=2p=1$
We get ${\lambda }_D=\frac{2^2\times 1^2}{R\left(2^2-1^2\right)}=1.33R$
Thus minimum wavelength is emitted when a transition takes place from $n=2$ to $p=1$ orbit.