Question

The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio $1:4$. The energy difference between them may be

• A. either $12.09eV$ or $3.4eV$
• B. either $2.55eV$ or $10.2eV$
• C. either $13.6eV$ or $3.4eV$
• D. either $3.4eV$ or $0.85eV$

Answer 1

Answer: (B)

either $2.55eV$ or $10.2eV$

$Explanation:$

$r\propto n^2$

where ,

$r=Radii$

$n=Energylevel$

$\frac{r_{n_2}}{r_{n_1}}=\frac{n_2^2}{n_1^2}=\frac{4}{1}$

$\frac{n_2}{n_1}=\frac{2}{1}$

$\therefore either{n}_1=1and{n}_2=2or{n}_1=2and{n}_2=4.$

From Energy equation:

$\Delta E=-13.6[\frac{1}{n_2^2}-\frac{1}{n_1^2}]eV$

$1)For{n}_1=1and{n}_2=2$

$\Delta E=-13.6[\frac{1}{2^2}-\frac{1}{1^2}]eV$

$\Delta E=10.2eV$

$2)For{n}_1=2and{n}_2=4$

$\Delta E=-13.6[\frac{1}{4^2}-\frac{1}{2^2}]eV$

$\Delta E=2.55eV$

$\therefore \Delta E=10.2eVor2.55eV$

$Hencethecorrectanswerisoption\left(B\right).$