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