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}

CBSEMCQ· 1mImportance★★★★★

51% · 42/83 Questions

✓ Free question

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.

Showing the 12 most recent of 30 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. If the wavelength of a photon is halved, then its frequency will become ______.$
›Reveal solutionSolution

Since nu = c/lambda for a photon, halving lambda directly doubles nu (c is a universal constant).

A photon's frequency and wavelength are related by nu = c/lambda, where c (speed of light) is fixed. If lambda is halved (lambda -> lambda/2), then nu = c/(lambda/2) = 2(c/lambda), i.e. the frequency becomes twice its original value.

✓Final answer
$doubled (2x the original frequency).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The mass of a photon is:$
(a) h/v
(b) hc/λ
(c) h/λ
(d) hν/c²
›Reveal solutionSolution

A photon's energy is $E=h\nu$ ; equating this to $E=mc^2$ gives its (relativistic/effective) mass $m=h\nu/c^2$ .

A photon has zero rest mass but carries energy $E=h\nu$ (Planck's relation) and momentum $p=h/\lambda=E/c$ . Using Einstein's mass-energy equivalence $E=mc^2$ for the energy it carries while in motion, its effective mass is $m=\dfrac{E}{c^2}=\dfrac{h\nu}{c^2}$ . The other options are quantities in disguise: $hc/\lambda=h\nu$ is the photon's energy, not its mass, and $h/\lambda$ is its momentum ( $p=h/\lambda$ ), not its mass.

✓Final answer
$h\nu/c^2$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The rest mass of photon is ______.$
›Reveal solutionSolution

A photon's rest mass is zero.

A photon is a quantum of electromagnetic radiation that always travels at the speed of light c in vacuum. According to relativity, any particle moving at speed c must have zero rest mass; otherwise its energy would be infinite. A photon does have energy (E = hν) and momentum (p = hν/c), but its rest mass (mass measured when at rest) is zero.

✓Final answer
$zero.$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The value of Planck's constant is ______.$
›Reveal solutionSolution

Planck's constant h ≈ 6.63 × 10⁻³⁴ J·s.

Planck's constant h relates the energy of a photon to its frequency by E = hν. Its accepted value is

h = 6.63 × 10⁻³⁴ joule-second (J·s).

It is one of the fundamental constants of nature and appears throughout quantum physics.

✓Final answer
$6.63 \times 10⁻³⁴ J\cdots.$
CBSE 2025Set 55/4/11 markMCQ
$Q. A beam of red light and a beam of blue light have equal intensities. Which of the following statements is true? (A) The blue beam has more number of photons than the red beam. (B) The red beam has more number of photons than the blue beam. (C) Wavelength of red light is lesser than the wavelength of blue light. (D) The blue light beam has lesser energy per photon than that in the red light beam.$
›Reveal solutionSolution
Since blue photons carry more energy than red photons, equal-intensity beams require more red photons to match the same total power. The red beam has more photons.

The key to this problem lies in understanding what "intensity" means and how photon energy depends on wavelength.

Intensity measures the power (energy per unit time) delivered per unit area. When two beams have equal intensities, they carry the same total energy per second through the same cross-sectional area, regardless of color.

Each photon carries energy $E = h\nu = \frac{hc}{\lambda}$ , where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Blue light has a shorter wavelength than red light ( $\lambda_{\text{blue}} < \lambda_{\text{red}}$ ), which means blue photons are individually more energetic than red photons.

If the total power delivered by both beams is the same, but blue photons pack more energy each, then fewer blue photons are needed to deliver that power. Conversely, more red photons are required to compensate for their lower individual energy.

Let me work through this quantitatively:

Express intensity in terms of photon count. If $n$ photons pass through area $A$ in time $t$ , the intensity is:

$I = \frac{\text{Total energy}}{A \cdot t} = \frac{n \cdot E_{\text{photon}}}{A \cdot t} = \frac{n \cdot hc/\lambda}{A \cdot t}$

Set up the equal-intensity condition. For red and blue beams with equal intensities:

$I_{\text{red}} = I_{\text{blue}}$ $\frac{n_{\text{red}} \cdot hc/\lambda_{\text{red}}}{A \cdot t} = \frac{n_{\text{blue}} \cdot hc/\lambda_{\text{blue}}}{A \cdot t}$

Simplify to find the photon ratio:

$n_{\text{red}} \cdot \frac{1}{\lambda_{\text{red}}} = n_{\text{blue}} \cdot \frac{1}{\lambda_{\text{blue}}}$ $\frac{n_{\text{red}}}{n_{\text{blue}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{blue}}}$

Apply the wavelength relationship. Since red light has a longer wavelength than blue light ( $\lambda_{\text{red}} > \lambda_{\text{blue}}$ ):

$\frac{n_{\text{red}}}{n_{\text{blue}}} > 1 \implies n_{\text{red}} > n_{\text{blue}}$
Now let's check each option:

(A) Claims blue has more photons — false, we just showed the opposite.

(B) Claims red has more photons — true, matches our derivation.

(C) Claims red wavelength is less than blue — false, red has longer wavelength.

(D) Claims blue photons have less energy — false, $E_{\text{blue}} = hc/\lambda_{\text{blue}} > hc/\lambda_{\text{red}} = E_{\text{red}}$ .

Tip

A quick mnemonic: "Lower energy photons need higher numbers" — to match the same total power, the beam with less energetic photons must have more of them.

✓Final answer
$The correct option is (B) : the red beam has more photons than the blue beam.$
CBSE 2025Set 55/5/11 markMCQ
$Q. Which of the following electromagnetic waves has photons of the largest momentum? (A) X-rays (B) AM radio waves (C) Microwaves (D) TV waves$
›Reveal solutionSolution
Photon momentum is $p = \frac{h}{\lambda}$ , so the wave with the shortest wavelength has the largest momentum. Among the options, X-rays have the shortest wavelength, hence the largest photon momentum.

Concept & Intuition

The momentum of a photon is not like the momentum of a massive particle. For a photon, momentum is purely a wave property, given by the de Broglie relation:
$p = \frac{h}{\lambda}$
where $h$ is Planck’s constant and $\lambda$ is the wavelength. This means: shorter wavelength → larger momentum. There is no dependence on amplitude or intensity — only wavelength matters.

So the question reduces to: which of these electromagnetic waves has the shortest wavelength? Let’s recall the electromagnetic spectrum order from longest to shortest wavelength:

Radio waves (including AM and TV) — longest wavelengths (metres to kilometres)

Microwaves — centimetres to millimetres

Infrared — micrometres

Visible light — hundreds of nanometres

Ultraviolet — tens of nanometres

X-rays — picometres to nanometres

Gamma rays — sub-picometre

Watch out

A common mistake is to think that higher frequency means higher energy (true), but then incorrectly assume that momentum depends on something else like the wave’s “penetrating power” or “ionising ability”. Stick to $p = h/\lambda$ — it’s the only formula that matters here.

Step-by-step solution

Write the momentum formula

For any photon, $p = \frac{h}{\lambda}$ . Since $h$ is constant, $p \propto \frac{1}{\lambda}$ .

Identify the wavelengths of each option

AM radio waves: wavelength $\approx 100$  m to $1000$  m (longest)

TV waves: wavelength $\approx 0.1$  m to $10$  m (still radio band)

Microwaves: wavelength $\approx 1$  mm to $30$  cm

X-rays: wavelength $\approx 0.01$  nm to $10$  nm (shortest among these)

Compare

Since $p \propto 1/\lambda$ , the smallest $\lambda$ gives the largest $p$ . X-rays have the smallest wavelength by many orders of magnitude.

Conclude

X-ray photons carry the largest momentum.

Tip

You don’t need to memorise exact numbers — just remember the order of the EM spectrum from longest to shortest wavelength: Radio → Microwave → Infrared → Visible → UV → X-ray → Gamma. The one furthest to the right among the options wins.

✓Final answer
$The correct option is (A) X-rays .$
CBSE 2025Set D1 markMCQ
$Q. What is the energy of a photon with a wavelength of 500 nm? ( Use c = 3 \times 10^8 m/s and h = 6.626 \times 10^-34 Js ) (A) 4 \times 10^-19 J (B) 2.5 \times 10^-19 J (C) 1.2 \times 10^-18 J (D) 6.6 \times 10^-19 J$
›Reveal solutionSolution

Photon energy E = hc/λ ≈ 4 × 10⁻¹⁹ J for λ = 500 nm.

The energy of a photon is

$E = \frac{hc}{\lambda}$

Substitute h = 6.626×10⁻³⁴ J·s, c = 3×10⁸ m/s, λ = 500 nm = 500×10⁻⁹ m:

$E = \frac{(6.626\times10^{-34})(3\times10^{8})}{500\times10^{-9}}$

$E = \frac{1.9878\times10^{-25}}{5\times10^{-7}} = 3.98\times10^{-19}\ \text{J}$

This rounds to 4 × 10⁻¹⁹ J.

✓Final answer
$(A) 4 \times 10⁻¹⁹ J.$
CBSE 2025Set A1 mark
$Q. Match Column 'A' item 'Frequency of light' with the correct option from Column 'B' and write the correct pair. Column 'B' options:$
(i) Minimum energy to emit electrons from the surface
(ii) Minimum frequency to emit electrons from the surface
(iii) Frequency of photon
(iv) Number of photons
(v) Moving particle
(vi) Photon
(vii) Einstein.
›Reveal solutionSolution

Frequency of light corresponds to option (iii): frequency of photon.

In the photon (particle) picture of light proposed by Einstein, a beam of light of frequency ν is regarded as a stream of photons, each carrying energy E = hν — so the 'frequency of light' (a wave concept) and the 'frequency of the photon' (used to compute each photon's quantum of energy) are simply the same physical quantity viewed from the two complementary (wave/particle) descriptions of light. Hence 'Frequency of light' matches '(iii) Frequency of photon'.

✓Final answer
$Frequency of light \to (iii) Frequency of photon.$
CBSE 2025Set ANNUAL1 mark
$Q. Electron volt (eV) is the unit of ................. (fill in the blank)$
›Reveal solutionSolution

The electron volt is a convenient small unit of energy, widely used in atomic and nuclear physics.

One electron volt is defined as the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt:

$1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$

Because atomic, photon, and nuclear energies are typically tiny fractions of a joule, the eV (and its multiples keV, MeV) is the standard convenient energy unit in this domain.

✓Final answer
$Energy.$
CBSE 2025Set ANNUAL1 mark
$4500\overset{\circ}{A}$
›Reveal solutionSolution

Visible-light power = 8% of 150 W; divide by the energy of one photon at 4500 Å.

Power emitted as visible light $= 8\%$ of $150\,$ W $= 0.08\times150 = 12\,$ W.

Energy of one photon at $\lambda=4500\,\text{Å}=4.5\times10^{-7}\,$ m:

$E = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^8}{4.5\times10^{-7}} \approx 4.42\times10^{-19}\ \text{J}$

Number of photons emitted per second:

$n = \frac{P}{E} = \frac{12}{4.42\times10^{-19}} \approx 2.71\times10^{19}\ \text{photons/s}$

✓Final answer
$2.71\times10^{19}$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The momentum of a photon of energy h.nu is$
(i) h.nu
(ii) h.nu/c
(iii) h.nu.c
(iv) h/nu
›Reveal solutionSolution

Photon momentum p = E/c = h(nu)/c.

A photon of frequency $\nu$ carries energy $E = h\nu$ . Being a massless quantum that moves at the speed of light, its momentum is $p = E/c$ . Therefore $p = \frac{h\nu}{c}$ (equivalently $p = h/\lambda$ since $c = \nu\lambda$ ).

✓Final answer
$(ii) h.nu/c.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The momentum (p) of photon is -$
(a) h/λ
(b) λ/h
(c) hC/λ
(d) hλ
›Reveal solutionSolution

A photon's momentum follows from combining its energy E = hc/λ with the relativistic relation E = pc for a massless particle.

A photon of frequency ν has energy $E = h\nu = \dfrac{hc}{\lambda}$ (since $c = \nu\lambda$ ).

A photon is massless and travels at speed c, so by the relativistic energy-momentum relation for a massless particle, $E = pc$ . Equating the two expressions for E:

$pc = \dfrac{hc}{\lambda} \implies p = \dfrac{h}{\lambda}$

✓Final answer
$(a) h/λ.$

←Previous questionQ1 — A particle is dropped from a height H. The de Broglie wavelength of th…→Next questionQ3 — Consider a beam of electrons (each electron with energy E₀) incident o…

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

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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