What is the minimum energy that must be given to a H atom in ground state so that it can emit an $H_\gamma$ line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such $H_\gamma$ photon?

What is the minimum energy that must be given to a H atom in ground state so that it can emit an H_γ line in B

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H_\gamma

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Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

Light does not carry its energy in a continuous stream. It comes in discrete packets — like individual drops instead of a steady hose. Each packet is called a photon: the smallest possible unit of light of a given frequency. You cannot have half a photon; it is all or nothing.

The Intuition: Colour Determines Energy

Red light and blue light are both light, but they behave differently — blue light causes sunburn and drives chemical reactions more readily than red. The reason is that the colour (frequency) of light directly fixes how much energy each photon carries, and blue photons carry more energy than red photons.

The Planck–Einstein Relation

The energy $E$ of a single photon is directly proportional to its frequency $f$ , and inversely proportional to its wavelength $\lambda$ :

E = h f = \frac{h c}{\lambda}

Where:

$E$ = energy of one photon (joules, J)
$f$ = frequency (hertz, Hz)
$\lambda$ = wavelength (metres, m)
$c$ = speed of light in vacuum ( $3.00 \times 10^8$ m/s)
$h$ = Planck's constant ( $6.626 \times 10^{-34}$ J·s)

Important

Planck's constant $h$ is extremely tiny, so an individual photon carries a very small amount of energy — which is why we never feel single photons striking our skin.

What This Tells You

Higher frequency = higher energy. Gamma rays have extremely energetic photons; radio waves have very low-energy photons.
Shorter wavelength = higher energy. Ultraviolet photons are more energetic than visible-light photons; infrared photons are less.
Energy is quantised. Light energy comes in fixed packets — you can have 1, 2, or 1000 photons, but never 0.5. This was Max Planck's revolutionary idea of 1900.

This is why UV light causes sunburn (UV photons carry enough energy to damage DNA, while visible photons do not), why X-ray photons are energetic enough to pass through soft tissue but get absorbed by bone, and why a photon needs enough energy to actually break a bond or drive a photochemical/biological reaction, rather than merely wavelength-dependent intensity.

A Concrete Example

Note

Question: Which has more energy — a photon of red light ( $\lambda = 700$ nm) or a photon of blue light ( $\lambda = 450$ nm)?

Since $E = \dfrac{hc}{\lambda}$ , energy is inversely proportional to wavelength, so the shorter-wavelength blue photon carries more energy:

Red: $E_{\text{red}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{700 \times 10^{-9}} \approx 2.84 \times 10^{-19}$ J …

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The formula $E = hf$ is not something you memorise and plug numbers into. It comes from a deep shift in how physicists understood light.

The problem that forced the formula

By the late 1800s, classical physics said light was a wave — continuous, carrying energy in proportion to its intensity (brightness). But blackbody radiation and the photoelectric effect showed something bizarre: when you shine light on a metal, electrons are ejected only if the light's frequency is above a certain threshold, no matter how bright the light is. Below that frequency, even the brightest lamp won't kick out a single electron. Classically this made no sense — a wave's energy depends on amplitude, not frequency.

Einstein's radical idea (1905)

Einstein proposed that light comes in discrete packets — photons — each carrying a fixed energy that depends only on its frequency, with Planck's constant $h$ as the proportionality:

A single photon's energy is $E = hf$ , where $f$ is the frequency of the light.

This directly explains the photoelectric effect: an electron needs a minimum energy (the work function $\phi$ ) to escape. If a photon's energy $hf < \phi$ , no electron is ejected — regardless of how many photons you send, because each photon interacts with one electron individually.

Important

$E = hf$ is not derived from deeper principles — it is a postulate justified by the experiments it explains, and it unified optics with quantum mechanics.

The equivalent forms

Since $c = f\lambda$ for light in vacuum, the same idea in terms of wavelength is:

E = \frac{hc}{\lambda}

Use this when you're given $\lambda$ instead of $f$ . A photon also carries momentum (despite having no mass): …

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$H_\gamma$ is the $n=5\to 2$ Balmer line, so the atom must be raised from $n=1$ to $n=5$ : minimum energy $=13.6\left(1-\tfrac{1}{25}\right)=13.06\,\text{eV}$ . By conservation of angular momentum the photon carries $L_5-L_2 = 3\hbar = \dfrac{3h}{2\pi}\approx 3.16\times10^{-34}\,\text{J s}$ .

1. Which transition is $H_\gamma$ ?

The Balmer series consists of transitions that terminate at $n=2$ . Counting from the longest wavelength: $H_\alpha = 3\to2$ , $H_\beta = 4\to2$ , and $H_\gamma = 5\to2$ . Therefore the electron must be present in the $n=5$ level for $H_\gamma$ to be emitted, and the atom starts in the ground state $n=1$ .

2. Minimum energy to make emission possible.

Using $E_n = -\dfrac{13.6}{n^2}\,\text{eV}$ , the energy needed to raise the atom from $n=1$ to $n=5$ is

\Delta E = E_5 - E_1 = -\frac{13.6}{25} - (-13.6) = 13.6\left(1 - \frac{1}{25}\right) = 13.6\times\frac{24}{25} = 13.06\,\text{eV}.

This is the minimum energy that must be supplied; once at $n=5$ , the atom can drop to $n=2$ and emit $H_\gamma$ .

3. Angular momentum of the $H_\gamma$ photon. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$Q. If an electron in the excited state falls to ground state, a photon of energy 5 eV is emitted, then the wavelength of the photon is nearly (A) 748 nm (B) 598 nm (C) 398 nm (D) 248 nm$
›Reveal solutionSolution
Tests the photon energy–wavelength relation for a hydrogen-atom-like transition; the emitted photon has $\lambda \approx 248\ \text{nm}$ .

Concept and Intuition

When an electron falls from an excited state to the ground state, the energy difference is carried away as a single photon, $E = h\nu = \dfrac{hc}{\lambda}$ . It is convenient to remember the handy constant $hc \approx 1240\ \text{eV·nm}$ , so that $E(\text{eV})\,\lambda(\text{nm}) = 1240$ .

Step-by-Step Solution

Given photon energy $E = 5\ \text{eV}$ .

Using $\lambda = \dfrac{1240}{E(\text{eV})}\ \text{nm}$ : $\lambda = \frac{1240}{5} = 248\ \text{nm}$ …
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AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$E_A < E_B < E_C$
›Reveal solutionSolution
Since the energy released in going directly from C to A must equal the energy released going C→B→A, the photon energies (not wavelengths) add, giving $\lambda_3=\lambda_1\lambda_2/(\lambda_1+\lambda_2)$ .

Concept and Intuition

Energy levels obey conservation of energy: the total energy dropped going from level C to level A is a fixed number, $E_C-E_A$ , regardless of the path taken. Whether the atom emits one photon directly (C→A) or two photons via an intermediate level (C→B then B→A), the energies released must add up to the same total. Photon energy is $E=hc/\lambda$ , so it is the reciprocals of the wavelengths that add, not the wavelengths themselves.

Step-by-Step Solution

Energy released in transition C→B: $E_1 = \dfrac{hc}{\lambda_1}$ .

Energy released in transition B→A: $E_2 = \dfrac{hc}{\lambda_2}$ .

Energy released in the direct transition C→A: $E_3 = \dfrac{hc}{\lambda_3}$ .

Conservation of energy: $E_3 = E_1+E_2$ , since $E_C-E_A = (E_C-E_B)+(E_B-E_A)$ .

So $\dfrac{hc}{\lambda_3} = \dfrac{hc}{\lambda_1}+\dfrac{hc}{\lambda_2} \Rightarrow \dfrac{1}{\lambda_3}=\dfrac{1}{\lambda_1}+\dfrac{1}{\lambda_2} = \dfrac{\lambda_1+\lambda_2}{\lambda_1\lambda_2}$ .

Invert: $\lambda_3 = \dfrac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}$ .

Common Mistakes …
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AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
$Q. The particle having zero mass is (A) proton (B) neutron (C) photon (D) electron$
›Reveal solutionSolution

The photon is the only listed particle with zero rest mass; protons, neutrons, and electrons are all massive particles.

Concept and Intuition

Particles are classified by whether they have rest mass. Matter particles like the proton, neutron, and electron all have well-defined nonzero rest masses (measured precisely in atomic mass units / MeV). The photon, the quantum of the electromagnetic field, has zero rest mass — it can never be brought to rest and always moves at speed $c$ in vacuum; its total energy is purely kinetic in the relativistic sense, given by $E=pc$ (the massless limit of $E^2=(pc)^2+(mc^2)^2$ ).

Step-by-Step Solution …
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Q.What is the minimum energy that must be given to a H atom in ground state so that it can emit an $H_\gamma$ line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such $H_\gamma$ photon?

Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

The Intuition: Colour Determines Energy

The Planck–Einstein Relation

What This Tells You

A Concrete Example

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The problem that forced the formula

Einstein's radical idea (1905)

The equivalent forms

Method: Series-Identification Recipe + Ladder-Sum Cross-Check

Steps

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution …

Q.What is the minimum energy that must be given to a H atom in ground state so that it can emit an HγH_\gammaHγ​ line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such HγH_\gammaHγ​ photon?

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Q.What is the minimum energy that must be given to a H atom in ground state so that it can emit an $H_\gamma$ line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such $H_\gamma$ photon?

Unlock everything free for 14 days