Q.A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n=4 level. Determine the wavelength and frequency of photon.
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🔒 Start your 14-day free trial to unlock the full solution →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 λ:
E=hf=λhc
Where:
- E = energy of one photon (joules, J)
- f = frequency (hertz, Hz)
- λ = wavelength (metres, m)
- c = speed of light in vacuum (3.00×108 m/s)
- h = Planck's constant (6.626×10−34 J·s)
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
Question: Which has more energy — a photon of red light (λ=700 nm) or a photon of blue light (λ=450 nm)?
Since E=λhc, energy is inversely proportional to wavelength, so the shorter-wavelength blue photon carries more energy:
- Red: Ered=700×10−9(6.626×10−34)(3.00×108)≈2.84×10−19 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 ϕ) to escape. If a photon's energy hf<ϕ, no electron is ejected — regardless of how many photons you send, because each photon interacts with one electron individually.
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=fλ for light in vacuum, the same idea in terms of wavelength is:
E=λhc
Use this when you're given λ instead of f. A photon also carries momentum (despite having no mass): …
E1=−13.6 eV,E4=−1613.6=−0.85 eV⟹ΔE=12.75 eV
Using hc=1240 eV⋅nm: λ=12.751240≈97.3 nm. …
The photon must supply exactly the energy gap between n=1 and n=4, which is 12.75 eV — giving a wavelength of 97.3 nm and a frequency of 3.08×1015 Hz.
Energy of the transition
For hydrogen, En=−n213.6 eV.
E1=−13.6 eV,E4=−1613.6=−0.85 eV
Since the atom is excited from n=1 to n=4, the photon must supply exactly this energy gap:
ΔE=E4−E1=(−0.85)−(−13.6)=12.75 eV
Converting to frequency
Convert to joules: ΔE=12.75×1.602×10−19=2.043×10−18 J.
E=hf⟹f=hE
f=6.626×10−342.043×10−18≈3.08×1015 Hz
(This lies in the ultraviolet range.)
Converting to wavelength
λ=fc=3.08×10153.00×108≈9.73×10−8 m=97.3 nm …
Method: Energy-Level Transition (Bohr Model)
This problem uses the Bohr model of the hydrogen atom, where the energy of an electron in level n is given by:
En=−n213.6 eV
The photon energy absorbed equals the difference between the final and initial energy levels.
Step 1: Write the energy of each level
For the ground state (n=1):
E1=−1213.6=−13.6 eV
For the excited state (n=4):
E4=−4213.6=−1613.6=−0.85 eV
Step 2: Find the photon energy absorbed
The photon must supply exactly the energy gap:
ΔE=E4−E1=(−0.85)−(−13.6)=12.75 eV
The negative signs cancel — the photon energy is positive, as expected for absorption.
Step 3: Convert photon energy to joules
For calculations with wavelength and frequency, use SI units:
1 eV=1.602×10−19 J
ΔE=12.75×1.602×10−19=2.04×10−18 J
Step 4: Find the frequency
Use Planck's relation E=hf:
f=hE=6.626×10−342.04×10−18
f=3.08×1015 Hz
--- …
Common Mistakes on This Photon Energy Problem
Students often lose marks on this seemingly straightforward question. Here are the most frequent errors and how to avoid each.
Mistake 1: Using the wrong energy formula
Many students plug E=hf directly without connecting it to the hydrogen energy levels. They forget that the photon's energy must exactly equal the difference between two atomic energy levels.
How to avoid: Always start by writing the energy of the n-th level in hydrogen:
En=−n213.6 eV
The photon energy is the gap between the final and initial states:
ΔE=E4−E1
Never use En alone — the photon doesn't "carry" the full energy of a level; it carries only the transition energy.
A common trap: students compute E4 and call that the photon energy. That gives a completely wrong answer — off by a factor of about 16.
Mistake 2: Sign errors in the energy difference
Students sometimes write ΔE=E1−E4 (reversed order) and get a negative value. Then they either panic or take the absolute value without understanding why.
How to avoid: The atom absorbs energy, so the final state has higher (less negative) energy. The correct order is:
ΔE=Efinal−Einitial=E4−E1
Since E4>E1 (both negative, but E4 is closer to zero), ΔE comes out positive — as it must for absorption.
For absorption: final minus initial. For emission: initial minus final. This keeps ΔE positive in both cases.
Mistake 3: Forgetting to convert eV to joules
The Rydberg constant is often given in eV, but h and c in the wavelength formula use SI units. Students plug eV directly into E=hc/λ and get nonsense.
How to avoid: After finding ΔE in eV, convert to joules:
1 eV=1.602×10−19 J
Then use:
λ=ΔEhcandf=hΔE
Where h=6.626×10−34 J⋅s and c=3.00×108 m/s.
Some exam problems let you use the shortcut: hc=1240 eV⋅nm. If you use this, ΔE stays in eV and λ comes out directly in nm. Check whether your exam allows this.
Mistake 4: Confusing wavelength and frequency formulas
Students mix up f=c/λ and λ=c/f, or use E=hc/λ but forget which variable to solve for.
How to avoid: Write the two relations clearly:
E=hf⇒f=hE
c=fλ⇒λ=fc=Ehc
Solve step by step: first find f from E, then find λ from f. Don't try to jump to both at once.
Mistake 5: Arithmetic errors with fractions
When computing E4−E1, students often mishandle the fractions:
E4=−1613.6=−0.85 eV
E1=−13.6 eV
ΔE=(−0.85)−(−13.6)=−0.85+13.6=12.75 eV …
- KCET 2024Set D-21 markMCQQ.The ratio of area of first excited state to ground state of orbit of hydrogen atom is (A) 1:16 (B) 1:4 (C) 4:1 (D) 16:1
›Reveal solutionSolution
r∝n2⇒ area ∝n4; for n=2 vs n=1 that is 24:14=16:1.
Step 1 — Bohr's radius formula
For a hydrogen-like atom, Bohr's quantisation of angular momentum (mvr=nℏ) combined with the Coulomb-force–centripetal-force balance gives
rn=πme2Zn2h2ε0⟹rn∝n2 (for fixed Z)
For hydrogen (Z=1), rn=n2a0 with a0=0.529 A˚.
Step 2 — From radius to area
The orbit is a circle, so the area enclosed is
An=πrn2
Since rn∝n2,
An∝(n2)2=n4
This n4 (not n2) is the crux of the question — the squaring of an already-squared quantity.
Step 3 — Identify the two states
- Ground state: n=1.
- First excited state: the next level up, n=2. (Not n=3 — the "first excited" state is the first level above the ground state.)
Step 4 — Take the ratio …
- COMEDK 2024Set 2024-M1 markMCQQ.The difference in energy levels of an electron at two excited levels is 13.75 eV. If it makes a transition from the higher energy level to the lower energy level then what will be the wave length of the emitted radiation? [given h=6.6×10−34 m2 kg s−1;c=3×108 ms−1;1 eV=1.6×10−19 J] (A) 900 nm (B) 90 A (C) 9000 nm (D) 900∘A
›Reveal solutionSolution
Using E=hc/λ, the 13.75 eV energy difference corresponds to a wavelength of 90 nm, i.e. 900 Å — matching option (D).
Step-by-step reasoning
- Convert the energy to joules.
E=13.75 eV×1.6×10−19 J/eV=2.2×10−18 J
- Solve for wavelength.
λ=Ehc=2.2×10−18(6.6×10−34)(3×108)=2.2×10−1819.8×10−26=9.0×10−8 m
- Convert to convenient units.
9.0×10−8 m=90 nm=900 A˚(since 1 nm=10 A˚)
This matches the option listing 900 Å. …
- KCET 2022Set B-31 markMCQQ.The radius of hydrogen atom in the ground state is 0.53 A∘. After collision with an electron, it is found to have a radius of 2.12 A∘, the principle quantum number 'n' of the final state of the atom is (A) n = 3 (B) n = 4 (C) n = 1 (D) n = 2
›Reveal solutionSolution
Bohr radii go as n2; the radius has grown by a factor of 4, so n2=4 and the atom is excited to n=2.
1. The Bohr radius law
Quantising the angular momentum (mvr=nh/2π) and balancing the Coulomb force against the centripetal requirement gives, for a hydrogen-like atom,
rn=πme2Zn2h2ε0=Zn2a0
where a0=0.53 A˚ is the Bohr radius. For hydrogen Z=1, so simply
rn=n2a0
The radius grows as the square of the principal quantum number — that quadratic dependence is the whole content of the problem.
2. Set up the ratio
Ground state: r1=0.53 A˚ (with n=1, consistent with the formula).
Final state after the collision: rn=2.12 A˚.
Taking the ratio kills a0 entirely:
r1rn=12a0n2a0=n2
3. Solve
n2=0.532.12=4
n=4=2
4. Physical reading …
- KCET 2019Set A-11 markMCQQ.Frequency of revolution of an electron revolving in nth orbit of H-atom is proportional to (A) n21 (B) n (C) n independent of n (D) n31
›Reveal solutionSolution
The frequency of revolution of an electron in the nth orbit of a hydrogen atom is proportional to n31, making option (D) correct.
The key here is to connect the frequency of revolution — how many times per second the electron circles the nucleus — to the orbital radius and velocity. In Bohr's model, the electron moves in a circular orbit under electrostatic attraction, and its angular momentum is quantized. Frequency is simply v/(2πr), so if we find how v and r depend on n, we can combine them.
Let's work through it step by step.
- Write the force balance for a stable orbit. The centripetal force is provided by the Coulomb attraction between the electron and the proton:
rmv2=r2ke2
where m is the electron mass, v its speed, r the orbit radius, k=1/(4πϵ0), and e the elementary charge.
- Apply Bohr's quantization condition. Angular momentum is an integer multiple of ℏ=h/(2π):
mvr=nℏ
- Solve for r in terms of n. From the quantization condition, v=nℏ/(mr). Substitute into the force equation:
rm(mrnℏ)2=r2ke2
Simplify:
mr3n2ℏ2=r2ke2
Multiply both sides by r3:
mn2ℏ2=ke2r
So:
r=mke2n2ℏ2
This shows r∝n2.
- Find v in terms of n. From mvr=nℏ, we have v=nℏ/(mr). Substitute r∝n2:
v∝n2n=n1
So v∝1/n.
- Now compute the frequency of revolution. Frequency f is the number of orbits per second: f=2πrv …
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