Question

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10⁻¹⁶s. The frequency of revolution of the electron in its first excited state (in s⁻¹) is

• A. 6.2 × 10¹⁵
• B. 7.8 × 10¹⁴
• C. 1.6 × 10¹⁴
• D. 5.6 × 10¹²

Answer 1

The correct option is B

7.8 × 10¹⁴

Step 1 Given Data,

$T_1=1.6\times {10}^{-16}s$

Step 2 Formula Used,

Angular momentum of the electron can be written in terms of principle quantum number,

$L_n=\frac{nh}{2\pi }$

n is the principle quantum

$f_2=\frac{2{r_1}^2}{{r^2}_2{T}_1}$

$f_2$is the frequency of revolution of the electron in its first excited state.

$r_1$is the radius of orbit in ground state.

$r_2$ is the radius of orbit in first excited state.

$T_1$is the period of revolution of electron in its ground state orbit.

Step 3 Calculating the frequency of revolution of the electron in its first excited state,

For the ground state, n=1

$L_1=\frac{h}{2\pi }=m{v}_1{r}_1$

For the first excited state, n=2

$$\begin{gathered} L_2=\frac{2h}{2\pi }=m{v}_2{r}_2 \\ {L}_2=2\left(\frac{h}{2\pi }\right)=m{v}_2{r}_2 \\ Therefore,L_2=2L_1 \end{gathered}$$

$L_2$ is the angular momentum of the electron in it's first excited state.

$L_1$ is the angular momentum of the electron in it's ground state.

$r_1$is the radius of orbit in ground state.

$r_2$ is the radius of orbit in first excited state.

$v_2$ is the velocity of orbit in first excited state.

$v_1$is the velocity of orbit in ground state.

$w_1$is the angular frequency of orbit in ground state.

$w_2$is the angular frequency of orbit in excited state.

$T_1$is the time period of revolution of electron in its ground state orbit

$$\begin{gathered} m{v}_2{r}_2=2m{v}_1{r}_1 \\ {v}_2{r}_2=2v_1{r}_1 \\ {w}_2{r_2}^2=2w_1{r_1}^2 \\ {f}_2=\frac{2f_1{r_1}^2}{{r_2}^2}=\frac{2{r_1}^2}{{T_1{r}_2}^2}v=w\times r,wcanbewriitenasf,f=\frac{1}{T} \end{gathered}$$

Now, the radius of orbit in a hydrogen is given by

$r=\frac{r_1{n}^2}{z}=r_1{n}^2$

For the first excited state,

$$\begin{gathered} r_2=r_1{n}^2 \\ {r}_2=r_1{2}^2 \\ {r}_2=4r_1 \\ \frac{r_2}{r_1}=4 \end{gathered}$$

Time period of revolution of electron in its ground state orbit in a hydrogen atom is
$T_1=1.6\times {10}^{-16}s$

$f_2=\frac{2{r_1}^2}{{r^2}_2{T}_1}=\frac{2}{16}\times \frac{1}{{10}^{-16}}=7.812\times {10}^{14}Hz$

Hence, option B is the correct answer.