JEE Main 2021

LEVELJEE Main

Animated Solution for Physics - Atoms and Nuclei: According to Bohr atom model, in which of the following transitions will the frequency be maximum ?

Select Answer:

Visualized Solution

$Energy Levels in Bohr Model$

$E_{n} = - \frac{13.6}{n ^{2}} eV$
$Δ E = E_{i} - E_{f}$

$Frequency of Emitted Photon$

$h ν = Δ E = 13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) eV$
$ν \propto (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})$

$Transition (a): n = 4 \to n = 3$

$ν_{a} \propto (\frac{1}{3 ^{2}} - \frac{1}{4 ^{2}})$
$ν_{a} \propto (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} \approx 0.048$

$Transition (b): n = 2 \to n = 1$

$ν_{b} \propto (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}})$
$ν_{b} \propto (1 - \frac{1}{4}) = \frac{3}{4} = 0.75$

$Transition (c): n = 5 \to n = 4$

$ν_{c} \propto (\frac{1}{4 ^{2}} - \frac{1}{5 ^{2}})$
$ν_{c} \propto (\frac{1}{16} - \frac{1}{25}) = \frac{9}{400} \approx 0.0225$

$Transition (d): n = 3 \to n = 2$

$ν_{d} \propto (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}})$
$ν_{d} \propto (\frac{1}{4} - \frac{1}{9}) = \frac{5}{36} \approx 0.138$

$Conclusion$

$ν_{b} > ν_{d} > ν_{a} > ν_{c}$
$Maximum frequency is for n = 2 \to n = 1$

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

Solution Diagram

The Ladder of Energy Levels

Imagine you are climbing a ladder, but this is no ordinary ladder. As you climb higher, the steps get closer and closer together. This is exactly how the energy levels of a hydrogen atom behave in the Bohr model!

The energy of an electron in the

n

-th orbit is given by the famous formula

E_{n} = - \frac{13.6}{n ^{2}} eV

. The negative sign simply means the electron is bound to the nucleus. When an electron jumps from a higher energy level (

n_{i}

) to a lower energy level (

n_{f}

), it sheds its excess energy by emitting a photon.

The Master Equation

The energy of this emitted photon is exactly equal to the energy difference between the two levels:

Δ E = E_{i} - E_{f} = 13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) eV

We also know from Planck's quantum theory that the energy of a photon is directly proportional to its frequency ($E = h u$). Therefore, the frequency $ u$ is directly proportional to the bracketed term:

u \propto (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}})

To find the transition that emits the photon with the maximum frequency, we just need to find which of the given transitions has the largest value for this bracketed term. Let's test them out!

Analyzing the Transitions

Option (a): $n = 4 \to n = 3$ Substituting the values, we get:

u_{a} \propto (\frac{1}{3 ^{2}} - \frac{1}{4 ^{2}}) = \frac{1}{9} - \frac{1}{16} = \frac{7}{144} \approx 0.048

Option (b): $n = 2 \to n = 1$ This is a jump to the ground state (Lyman series):

u_{b} \propto (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = 1 - \frac{1}{4} = 0.75

Option (c): $n = 5 \to n = 4$ These levels are far from the nucleus and very close to each other:

u_{c} \propto (\frac{1}{4 ^{2}} - \frac{1}{5 ^{2}}) = \frac{1}{16} - \frac{1}{25} = \frac{9}{400} \approx 0.0225

Option (d): $n = 3 \to n = 2$ This is the first line of the visible Balmer series:

u_{d} \propto (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \approx 0.138

The Visual Intuition

Comparing the values,

0.75

is massively larger than the rest. So, the transition from

n = 2

to

n = 1

produces the highest frequency.

But wait, there is a beautiful shortcut here! If you look at the energy values themselves: -

E_{1} = - 13.6 eV

-

E_{2} = - 3.4 eV

-

E_{3} = - 1.51 eV

The gap between

n = 1

and

n = 2

is a whopping $10.2 eV$ . The entire remaining energy from

n = 2

all the way to infinity (

n = \infty

) is only

3.4 eV

!

This means that any transition that ends at the ground state (

n = 1

) will always release more energy—and thus a higher frequency photon—than any transition that happens entirely among the higher excited states.

Final Answer: The maximum frequency is emitted during the transition

n = 2 \to n = 1

.

Common Traps & FAQs

Why is the frequency proportional to

(1/ n_{f}^{2} - 1/ n_{i}^{2})

?

Because the energy of a photon is

E = h ν

, and the energy difference between two Bohr orbits is

Δ E = 13.6 \times (1/ n_{f}^{2} - 1/ n_{i}^{2})

eV. Equating them gives the proportionality.

Can I solve this without calculating all values?

Absolutely! The energy gap between

n = 1

and

n = 2

is

10.2

eV. The total energy from

n = 2

to

n = \infty

is only

3.4

eV. So any transition ending at

n = 1

will always have a higher frequency than any transition that doesn't.

What if the question asked for maximum wavelength instead?

Wavelength is inversely proportional to frequency (

c = ν λ

). So, the transition with the *minimum* frequency (smallest energy gap) would have the maximum wavelength. In this case,

n = 5 \to n = 4

.

Does this apply to other hydrogen-like ions like He+ or Li2+?

Yes, the proportionality remains exactly the same. The actual energy values would be multiplied by

Z^{2}

, but the relative comparison between transitions doesn't change.

Why are the energy levels negative?

The negative sign indicates that the electron is bound to the nucleus. An energy of

0

eV means the electron is completely free (ionized) and at rest at an infinite distance.

Similar Questions

Which of the following transitions in hydrogen atoms emit photons of highest frequency?

n = 2

n = 6

n = 6

n = 2

n = 2

n = 1

n = 1

n = 2

In a hydrogen atom, electron makes a transition from $(n + 1)$ th level to the $n$ th level. If $n >> 1$ , the frequency of radiation emitted is proportional to

\frac{1}{n}

\frac{1}{n ^{3}}

\frac{1}{n ^{2}}

\frac{1}{n ^{4}}

In a hydrogen like atom electron make transition from an energy level with quantum number $n$ to another with quantum number $(n - 1)$ . If $n >> 1$ , the frequency of radiation emitted is proportional to

\frac{1}{n}

\frac{1}{n ^{2}}

\frac{1}{n ^{3/2}}

\frac{1}{n ^{3}}

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

The transition from the state $n = 4$ to $n = 3$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

2 \to 1

3 \to 2

4 \to 2

5 \to 4

The transition from the state $n = 4$ to $n = 3$ in a hydrogen like atom results in ultraviolet radiation. Infrared radition will be obtained in the transition from

2 \to 1

3 \to 2

4 \to 2

5 \to 4

A particular hydrogen like ion emits radiation of frequency $2.92 \times 1 0^{15}$ Hz when it makes transition from $n = 3$ to $n = 1$ . The frequency in Hz of radiation emitted in transition from $n = 2$ to $n = 1$ will be

0.44 \times 1 0^{15}

6.57 \times 1 0^{15}

4.38 \times 1 0^{15}

2.46 \times 1 0^{15}

The electron in a hydrogen atom makes a transition $n_{1} \to n_{2}$ , where $n_{1}$ and $n_{2}$ are the principal quantum numbers of two states. Assume the Bohr's model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of $n_{1}$ and $n_{2}$ are

* Multiple Correct Options

n_{1} = 4, n_{2} = 2

n_{1} = 8, n_{2} = 2

n_{1} = 8, n_{2} = 1

n_{1} = 6, n_{2} = 3

LEVELJEE Advanced

If the atom $_{100} Fm^{257}$ follows the Bohr's model and the radius of last orbit of $_{100} Fm^{257}$ is $n$ times the Bohr radius, then find $n$

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $λ$ . If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be

\frac{27}{20} λ

\frac{25}{16} λ

\frac{20}{27} λ

\frac{16}{25} λ