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)

n = 2 to p = 1

Wavelength of light emitted due to transition from $n$ to $p$ orbits $\frac{1}{\lambda }=R(\frac{1}{p^2}-\frac{1}{n^2})$

$⟹$ $\lambda =\frac{n^2{p}^2}{R(n^2-p^2)}$

Case (A) : $n=5$ $p=4$

We get ${\lambda }_A=\frac{5^2\times 4^2}{R(5^2-4^2)}=44.44R$

Case (B) : $n=4$ $p=3$

We get ${\lambda }_B=\frac{4^2\times 3^2}{R(4^2-3^2)}=20.57R$

Case (C) : $n=3$ $p=2$

We get ${\lambda }_C=\frac{3^2\times 2^2}{R(3^2-2^2)}=7.2R$

Case (D) : $n=2$ $p=1$

We get ${\lambda }_D=\frac{2^2\times 1^2}{R(2^2-1^2)}=1.33R$

Thus minimum wavelength is emitted when a transition takes place from $n=2$ to $p=1$ orbit.