Question

Orbital frequency of electron in $n$th orbit of hydrogen is twice that of $2$nd orbit. Calculate the value of n.

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

Answer: (A)

$1$

Since we are considering a hydrogen atom, we can apply Bohr’s model. We know:

$\nu =\frac{{\nu }_0}{n^2}$

$r=r_0{n}^2$

where,

$\nu =$ Orbital velocity

$r=$Radius of the orbit of an electron.

${\nu }_0,r_0=$Constants

Thus the angular velocity of the electron is:

$\omega =\frac{\nu }{r}\propto \frac{1}{n^3}$

The frequency of the electron is:

$\omega =\frac{\nu }{2\Pi }\propto \frac{1}{n^3}$

Thus,

$\frac{{\nu }_n}{{\nu }_2}=4\Rightarrow {\left(\frac{2}{n}\right)}^3=4\Rightarrow \left(\frac{8}{n^3}\right)=2\Rightarrow n^3=2\Rightarrow n=\sqrt[3]{32}\approx 1$

$\therefore n=1$

$Hencethecorrectanswerisoption\left(A\right).$