The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1\ \text{MeV}$ energy is nearly (a) $1.2\ \text{nm}$ (b) $1.2 \times 10^{-3}\ \text{nm}$ (c) $1.2 \times 10^{-6}\ \text{nm}$ (d) $1.2 \times 10^{1}\ \text{nm}$

> [!TLDR] > The photon must supply exactly the binding energy of the proton (1 MeV). Using $E = hc/\lambda$, the wavelength comes out to about $1.24 \times 10^{-3}\ \text{nm}$, which matches option **(B)**. The core idea here is that removing a proton from a nucleus requires overcoming the nuclear binding force. That binding energy is given as 1 MeV — the minimum energy a photon must carry to eject the proton. Since a photon’s energy is inversely proportional to its wavelength, we can directly compute the wavelength. A common pitfall is forgetting to convert units properly or mixing up the energy-wavelength relation for photons. Let’s walk through it cleanly. 1. **Recall the photon energy-wavelength relation** For any photon, $E = \frac{hc}{\lambda}$, where $h$ is Planck’s constant and $c$ is the speed of light. The product $hc$ is a very useful constant: $$hc = 1240\ \text{eV·nm}$$ (This is exact enough for all exam purposes — it comes from $h = 4.135667 \times 10^{-15}\ \text{eV·s}$ and $c = 2.998 \times 10^{8}\ \text{m/s}$, giving $hc \approx 1240\ \text{eV·nm}$.) 2. **Set the photon energy equal to the binding energy** The photon must have $E = 1\ \text{MeV} = 10^6\ \text{eV}$. So: $$\frac{hc}{\lambda} = 10^6\ \text{eV}$$ 3. **Solve for $\lambda$** $$\lambda = \frac{hc}{10^6\ \text{eV}} = \frac{1240\ \text{eV·nm}}{10^6\ \text{eV}} = 1.24 \times 10^{-3}\ \text{nm}$$ 4. **Match with the options** The value $1.24 \times 10^{-3}\ \text{nm}$ is extremely close to $1.2 \times 10^{-3}\ \text{nm}$ — the slight difference is due to rounding $hc$ to 1240 instead of 1239.84. In multiple-choice exams, this is the intended match. > [!WARNING] > A very common mistake is to use $E = hf$ and then forget that $c = f\lambda$, or to mix up units (e.g., using $hc = 1240\ \text{eV·nm}$ but then treating the energy in MeV without converting to eV). Always convert MeV to eV first: $1\ \text{MeV} = 10^6\ \text{eV}$. > [!TIP] > Memorising $hc =

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1MeV en

NCERT Exemplar · Q2

1\ \text{MeV}

(a)

1.2\ \text{nm}

(b)

1.2 \times 10^{-3}\ \text{nm}

(c)

1.2 \times 10^{-6}\ \text{nm}

(d)

1.2 \times 10^{1}\ \text{nm}

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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
Blue: $E_{\text{blue}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{450 \times 10^{-9}} \approx 4.42 \times 10^{-19}$ J

The blue photon carries about 1.5 times the energy of the red photon.

Watch out

A common mistake is to think brighter light means more energetic photons. Brightness is the number of photons per second, not the energy per photon. A bright red light has many low-energy photons; a dim blue light has fewer but higher-energy ones.

The Big Picture

Photon energy is the bridge between the wave nature of light (frequency, wavelength) and its particle nature (energy packets) — one of the foundational ideas of quantum mechanics: at the smallest scales, energy is not continuous but comes in discrete, indivisible units.

Photon energy, given by the Planck-Einstein relation E = hf, is one of the most fundamental formulas in the NCERT Class 12 Physics Dual Nature of Radiation and Matter chapter, and "photon energy formula and calculation" is a heavily searched query among students preparing for CBSE boards, JEE Main, and NEET. Because this idea links directly to the photoelectric effect and atomic spectra, it also anchors several "modern physics important questions" compiled for competitive-exam revision.

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):

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

This follows from special relativity: $E^2 = (pc)^2 + (mc^2)^2$ , and for a photon $m = 0$ , so $E = pc$ , hence $p = h/\lambda$ .

Note

$h = 6.626 \times 10^{-34} \ \text{J·s}$ (Planck's constant)
$c = 3.0 \times 10^8 \ \text{m/s}$ (speed of light)
For atomic-scale problems use electronvolts: $1 \ \text{eV} = 1.602 \times 10^{-19} \ \text{J}$

A common exam trap

$E = hf$ means higher frequency = more energy per photon — true. But a brighter light does not mean higher photon energy. Intensity is the number of photons per second per area; each photon still carries $hf$ .

Watch out

Do not confuse intensity (number of photons) with frequency (energy per photon). They are independent.

Final answer: $E = hf$

The photon must supply exactly the binding energy of the proton (1 MeV). Using $E = hc/\lambda$ , the wavelength comes out to about $1.24 \times 10^{-3}\ \text{nm}$ , which matches option (B).

The core idea here is that removing a proton from a nucleus requires overcoming the nuclear binding force. That binding energy is given as 1 MeV — the minimum energy a photon must carry to eject the proton. Since a photon’s energy is inversely proportional to its wavelength, we can directly compute the wavelength.

A common pitfall is forgetting to convert units properly or mixing up the energy-wavelength relation for photons. Let’s walk through it cleanly.

Recall the photon energy-wavelength relation For any photon, $E = \frac{hc}{\lambda}$ , where $h$ is Planck’s constant and $c$ is the speed of light. The product $hc$ is a very useful constant:

hc = 1240\ \text{eV·nm}

(This is exact enough for all exam purposes — it comes from $h = 4.135667 \times 10^{-15}\ \text{eV·s}$ and $c = 2.998 \times 10^{8}\ \text{m/s}$ , giving $hc \approx 1240\ \text{eV·nm}$ .)

Set the photon energy equal to the binding energy The photon must have $E = 1\ \text{MeV} = 10^6\ \text{eV}$ . So:

\frac{hc}{\lambda} = 10^6\ \text{eV}

Solve for $\lambda$

\lambda = \frac{hc}{10^6\ \text{eV}} = \frac{1240\ \text{eV·nm}}{10^6\ \text{eV}} = 1.24 \times 10^{-3}\ \text{nm}

Match with the options The value $1.24 \times 10^{-3}\ \text{nm}$ is extremely close to $1.2 \times 10^{-3}\ \text{nm}$ — the slight difference is due to rounding $hc$ to 1240 instead of 1239.84. In multiple-choice exams, this is the intended match.

Watch out

A very common mistake is to use $E = hf$ and then forget that $c = f\lambda$ , or to mix up units (e.g., using $hc = 1240\ \text{eV·nm}$ but then treating the energy in MeV without converting to eV). Always convert MeV to eV first: $1\ \text{MeV} = 10^6\ \text{eV}$ .

Tip

Memorising $hc = 1240\ \text{eV·nm}$ saves enormous time. For any photon energy in eV, the wavelength in nm is simply $1240/E$ . For MeV energies, just shift the decimal: $1240 / 10^6 = 1.24 \times 10^{-3}$ .

✓Final answer

1.2 \times 10^{-3}\ \text{nm}

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

KCET 2024Set D-21 markMCQ
$1:16$
›Reveal solutionSolution
$r\propto n^2 \Rightarrow$ area $\propto n^4$ ; for $n=2$ vs $n=1$ that is $2^4:1^4 = 16:1$ .

Step 1 — Bohr's radius formula

For a hydrogen-like atom, Bohr's quantisation of angular momentum ( $mvr = n\hbar$ ) combined with the Coulomb-force–centripetal-force balance gives
$r_n=\frac{n^{2}h^{2}\varepsilon_{0}}{\pi m e^{2}Z}\qquad\Longrightarrow\qquad r_n\propto n^{2}\ \ (\text{for fixed }Z)$
For hydrogen ( $Z=1$ ), $r_n = n^2 a_0$ with $a_0 = 0.529\ \text{Å}$ .

Step 2 — From radius to area

The orbit is a circle, so the area enclosed is
$A_n=\pi r_n^{2}$
Since $r_n \propto n^2$ ,
$A_n \propto \left(n^{2}\right)^{2} = n^{4}$
This $n^4$ (not $n^2$ ) 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
$\frac{A_{2}}{A_{1}}=\frac{\pi r_{2}^{2}}{\pi r_{1}^{2}}=\frac{(2^{2}a_0)^{2}}{(1^{2}a_0)^{2}}=\frac{(4a_0)^{2}}{(a_0)^{2}}=\frac{16a_0^{2}}{a_0^{2}}=16$
The question asks for first excited : ground, i.e.
$A_2 : A_1 = 16:1$
Step 5 — Guard against the traps

$4:1$ (option C) is the ratio of the radii, not the areas.

$1:4$ and $1:16$ have the ratio inverted — the excited orbit is bigger, so the ratio must be greater than 1.

✓Final answer
$16:1$

ANSWER: D
COMEDK 2024Set 2024-M1 markMCQ
$13.75 \mathrm{~eV}$
›Reveal solutionSolution
Using $E=hc/\lambda$ , 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.

$E = 13.75\ \text{eV} \times 1.6\times10^{-19}\ \text{J/eV} = 2.2\times10^{-18}\ \text{J}$

Solve for wavelength.

$\lambda = \frac{hc}{E} = \frac{(6.6\times10^{-34})(3\times10^8)}{2.2\times10^{-18}} = \frac{19.8\times10^{-26}}{2.2\times10^{-18}} = 9.0\times10^{-8}\ \text{m}$

Convert to convenient units.

$9.0\times10^{-8}\ \text{m} = 90\ \text{nm} = 900\ \text{Å} \quad (\text{since } 1\ \text{nm}=10\ \text{Å})$
This matches the option listing 900 Å.

Watch out

Don't confuse nm and Å — 1 nm equals 10 Å, not 100 Å. Here 90 nm equals 900 Å.

Tip

Quick shortcut: $E(\text{eV}) \approx \dfrac{1240}{\lambda(\text{nm})}$ , so $\lambda \approx 1240/13.75 \approx 90.2$ nm — confirming the result fast.

✓Final answer
$The correct option is (D) : 900 Å.$
KCET 2022Set B-31 markMCQ
$0.53~\overset{\circ}{A}$
›Reveal solutionSolution

Bohr radii go as $n^2$ ; the radius has grown by a factor of 4, so $n^2 = 4$ and the atom is excited to $n = 2$ .

1. The Bohr radius law

Quantising the angular momentum ( $mvr = nh/2\pi$ ) and balancing the Coulomb force against the centripetal requirement gives, for a hydrogen-like atom,
$r_n = \frac{n^{2}h^{2}\varepsilon_0}{\pi m e^{2} Z} \;=\; \frac{n^{2}}{Z}\,a_0$
where $a_0 = 0.53\ \text{Å}$ is the Bohr radius. For hydrogen $Z = 1$ , so simply
$\boxed{\,r_n = n^{2}\,a_0\,}$
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: $r_1 = 0.53\ \text{Å}$ (with $n=1$ , consistent with the formula).

Final state after the collision: $r_n = 2.12\ \text{Å}$ .

Taking the ratio kills $a_0$ entirely:
$\frac{r_n}{r_1} = \frac{n^{2}a_0}{1^{2}a_0} = n^{2}$
3. Solve
$n^{2} = \frac{2.12}{0.53} = 4$ $n = \sqrt{4} = 2$
4. Physical reading

The collision with the electron transferred just enough energy to lift the atom from the ground state to the first excited state. Check consistency with the energy ladder: $E_n = -13.6/n^2$ eV, so $E_2 - E_1 = -3.4 - (-13.6) = 10.2\ \text{eV}$ — precisely the well-known first excitation energy of hydrogen. Everything is self-consistent.

(Trap: option (C) $n=1$ would mean nothing happened, and $n=3$ or $4$ would require the radius to be $9a_0 = 4.77\ \text{Å}$ or $16a_0 = 8.48\ \text{Å}$ , not $2.12\ \text{Å}$ .)

✓Final answer
$The correct option is (D) — n = 2.$

ANSWER: D
KCET 2019Set A-11 markMCQ
$n^{th}$
›Reveal solutionSolution
The frequency of revolution of an electron in the $n^{th}$ orbit of a hydrogen atom is proportional to $\frac{1}{n^3}$ , 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 $v/(2\pi r)$ , 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:

$\frac{mv^2}{r} = \frac{ke^2}{r^2}$
where $m$ is the electron mass, $v$ its speed, $r$ the orbit radius, $k = 1/(4\pi\epsilon_0)$ , and $e$ the elementary charge.

Apply Bohr's quantization condition. Angular momentum is an integer multiple of $\hbar = h/(2\pi)$ :

$mvr = n\hbar$

Solve for $r$ in terms of $n$ . From the quantization condition, $v = n\hbar/(mr)$ . Substitute into the force equation:

$\frac{m}{r} \left( \frac{n\hbar}{mr} \right)^2 = \frac{ke^2}{r^2}$
Simplify:
$\frac{n^2 \hbar^2}{m r^3} = \frac{ke^2}{r^2}$
Multiply both sides by $r^3$ :
$\frac{n^2 \hbar^2}{m} = ke^2 r$
So:
$r = \frac{n^2 \hbar^2}{m k e^2}$
This shows $r \propto n^2$ .

Find $v$ in terms of $n$ . From $mvr = n\hbar$ , we have $v = n\hbar/(mr)$ . Substitute $r \propto n^2$ :

$v \propto \frac{n}{n^2} = \frac{1}{n}$
So $v \propto 1/n$ .

Now compute the frequency of revolution. Frequency $f$ is the number of orbits per second:

$f = \frac{v}{2\pi r}$
Substitute the proportionalities $v \propto 1/n$ and $r \propto n^2$ :
$f \propto \frac{1/n}{n^2} = \frac{1}{n^3}$
Watch out

A common mistake is to think frequency is proportional to $1/n^2$ because energy is proportional to $1/n^2$ . But frequency here is mechanical revolution frequency, not the frequency of emitted radiation (which relates to energy differences). They are different quantities.

Tip

You can also derive this directly from the known expressions: $r_n = n^2 a_0$ and $v_n = \alpha c / n$ , where $a_0$ is the Bohr radius and $\alpha$ the fine-structure constant. Then $f = v_n/(2\pi r_n) \propto (1/n)/(n^2) = 1/n^3$ .

✓Final answer
$\frac{1}{n^3}$

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Q.The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1\ \text{MeV}$ energy is nearly

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

The Big Picture

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

A common exam trap

Method: Converting a Threshold/Binding Energy into a Photon Wavelength

Steps

Step 1 — Bohr's radius formula

Step 2 — From radius to area

Step 3 — Identify the two states

Step 4 — Take the ratio

Step 5 — Guard against the traps

1. The Bohr radius law

2. Set up the ratio

3. Solve

4. Physical reading

Q.The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV1\ \text{MeV}1 MeV energy is nearly

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Q.The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1\ \text{MeV}$ energy is nearly

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