The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 x 10^-

Q: Why is the first excited state $n=2$ and not $n=1$?

The ground state is the lowest possible energy level, which is $n=1$. When the electron absorbs energy and jumps to the next available higher energy level, it reaches the 'first excited state', which corresponds to $n=2$.

Q: How did we get $T \propto n^3$?

In Bohr's model, the radius of the $n$th orbit is $r_n \propto n^2$, and the velocity of the electron is $v_n \propto \frac{1}{n}$. Since Time Period $T = \frac{\text{Distance}}{\text{Speed}} = \frac{2\pi r_n}{v_n}$, substituting the proportionalities gives $T \propto \frac{n^2}{1/n} = n^3$.

Q: Why do we need to calculate frequency instead of just stopping at time period?

Always read the final sentence of the question carefully! The question explicitly asks for the 'frequency of revolution', which is the reciprocal of the time period ($f = \frac{1}{T}$).

Q: Does this proportionality hold for other atoms like Helium?

For hydrogen-like ions (like $\text{He}^+$, $\text{Li}^{2+}$), you must also account for the atomic number $Z$. The radius is $r \propto \frac{n^2}{Z}$ and velocity is $v \propto \frac{Z}{n}$. Therefore, the time period becomes $T \propto \frac{n^3}{Z^2}$.

Q: What is the physical significance of $10^{14}$ Hz frequency?

A frequency of $10^{14}$ Hz means the electron completes $100,000,000,000,000$ (one hundred trillion) full orbits around the nucleus every single second! This mind-boggling speed is a hallmark of quantum mechanics at the atomic scale.

Question

The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6 	imes 10^{-16}$ s. The frequency of revolution of the electron in its first excited state (in Hz) is

Options:

(A) $1.6 	imes 10^{14}$

(B) $5.6 	imes 10^{12}$

(C) $6.2 	imes 10^{15}$

(D) $7.8 	imes 10^{14}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a hydrogen atom. In its ground state, the electron is zipping around the nucleus in the first orbit, where the principal quantum number is $n=1$. We are given its time period, $T_1 = 1.6 	imes 10^{-16} 	ext{ s}$, which is the time taken to complete one full revolution.

How do we mathematically define this time period? It is simply the total distance traveled in one orbit divided by the orbital velocity. Since the orbit is circular, the distance is the circumference $2\pi r_n$, and the velocity is $v_n$. Thus, the time period for the $n$th orbit is given by:

$$T_n = \frac{2\pi r_n}{v_n}$$

### The Master Proportionality
Now, from Bohr's model, we know how the radius and velocity scale with the principal quantum number $n$. The radius $r_n$ is directly proportional to the square of the orbit number ($r_n \propto n^2$), and the orbital velocity $v_n$ is inversely proportional to the orbit number ($v_n \propto \frac{1}{n}$).

Let's substitute these proportionalities into our time period formula. The numerator gets an $n^2$, and the denominator gets a $\frac{1}{n}$. When that $n$ flips up to the numerator, we find a beautiful relationship:

$$T_n \propto \frac{n^2}{1/n} \implies T_n \propto n^3$$

This means the time period of revolution is directly proportional to the cube of the principal quantum number!

### Calculating the New Time Period
The question asks about the **first excited state**, which corresponds to $n=2$. Since the time period scales as $n^3$, the new time period $T_2$ will be $2^3$, or $8$ times the ground state time period $T_1$.

$$T_2 = T_1 	imes 2^3 = 8 T_1$$

Let's plug in the given value of $T_1$. Multiplying $8$ by $1.6 	imes 10^{-16} 	ext{ s}$ gives us the time period for the second orbit:

$$T_2 = 8 	imes 1.6 	imes 10^{-16} = 12.8 	imes 10^{-16} 	ext{ s}$$

### Final Calculation: Finding the Frequency
But wait, there is a catch here. The question does not ask for the time period; it asks for the **frequency**. Frequency is simply the reciprocal of the time period. It tells us how many revolutions the electron makes in one second.

$$f_2 = \frac{1}{T_2}$$

Let's calculate it carefully:

$$f_2 = \frac{1}{12.8 	imes 10^{-16}} = \frac{1}{12.8} 	imes 10^{16}$$

Since $\frac{1}{12.8} = 0.078125$, we get:

$$f_2 = 0.078125 	imes 10^{16} 	ext{ Hz}$$

Adjusting the powers of ten to match standard scientific notation, we arrive at our final answer:

$$f_2 \approx 7.8 	imes 10^{14} 	ext{ Hz}$$

This perfectly matches option (d). Always remember to read the question carefully to see whether it asks for the time period or the frequency!

Animated Solution for Physics - Atoms and Nuclei: The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6 \times 1 0^{- 16}$ s. The frequency of revolution of the electron in its first excited state (in Hz) is

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Visualized Solution

$T_{1} = 1.6 \times 1 0^{- 16} s$

$T_{n} = \frac{2 π r _{n}}{v _{n}}$

$r_{n} \propto n^{2}, v_{n} \propto \frac{1}{n}$

$T_{n} \propto n^{3}$

$T_{2} = T_{1} \times 2^{3}$

$T_{2} = 8 \times 1.6 \times 1 0^{- 16} s$

$f_{2} = \frac{1}{T _{2}}$

$f_{2} = 7.8 \times 1 0^{14} Hz$

$T_{n} \propto \frac{n ^{3}}{Z ^{2}}$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Analyzing the Setup Imagine a hydrogen atom

The Master Proportionality Now, from Bohr's model, we know how the radius and velocity scale with the principal quantum number $n$

Calculating the New Time Period The question asks about the first excited state, which corresponds to $n = 2$

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​=1.6×10−16 s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​=vn​2πrn​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rn​∝n2,vn​∝n1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​∝n3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​=T1​×23

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​=8×1.6×10−16 s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​=T2​1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​=7.8×1014 Hz

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​∝Z2n3​