Question

Which transition emits light of lower frequency ?

• A. $n_1=6$ to $n_2=2$
• B. $n_1=2$ to $n_2=1$
• C. $n_1=2$ to $n_2=6$
• D. $n_1=1$ to $n_2=2$

Answer 1

Answer: (A)

From Bohr's concept, the wavelength $(\lambda )$ of emitted light is given by
$\frac{1}{\lambda }=R\left(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right)$
For $n_1=1,n_2=2$
$\therefore \frac{1}{{\lambda }_1}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
$=R\left(\frac{1}{1}-\frac{1}{4}\right)=0.75R$
For $n_1=2,n_2=6$
$\frac{1}{{\lambda }_2}=R\left(\frac{1}{4}-\frac{1}{36}\right)=0.22R$
Also frequency $(v)\propto \frac{1}{\lambda }$.
and $E=hv$
Hence, for minimum $E,n_2=6$ and $n_1=2$.