Skip to content
NCERT Exemplar · Q2

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} energy is nearly

(a) 1.2 nm1.2\ \text{nm}
(b) 1.2×10−3 nm1.2 \times 10^{-3}\ \text{nm}
(c) 1.2×10−6 nm1.2 \times 10^{-6}\ \text{nm}
(d) 1.2×101 nm1.2 \times 10^{1}\ \text{nm}
Gujarat GsebMCQ· 1mImportance★★★★★
51% · 42/83 Questions
✓ Free question

The photon must supply exactly the binding energy of the proton (1 MeV). Using E=hc/λE = hc/\lambda, the wavelength comes out to about 1.24×10−3 nm1.24 \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=hcλE = \frac{hc}{\lambda}, where hh is Planck’s constant and cc is the speed of light. The product hchc is a very useful constant:

hc=1240 eV⋅nmhc = 1240\ \text{eV·nm}

(This is exact enough for all exam purposes — it comes from h=4.135667×10−15 eV⋅sh = 4.135667 \times 10^{-15}\ \text{eV·s} and c=2.998×108 m/sc = 2.998 \times 10^{8}\ \text{m/s}, giving hc≈1240 eV⋅nmhc \approx 1240\ \text{eV·nm}.)

  1. Set the photon energy equal to the binding energy The photon must have E=1 MeV=106 eVE = 1\ \text{MeV} = 10^6\ \text{eV}. So:

hcλ=106 eV\frac{hc}{\lambda} = 10^6\ \text{eV}

  1. Solve for λ\lambda

λ=hc106 eV=1240 eV⋅nm106 eV=1.24×10−3 nm\lambda = \frac{hc}{10^6\ \text{eV}} = \frac{1240\ \text{eV·nm}}{10^6\ \text{eV}} = 1.24 \times 10^{-3}\ \text{nm}

  1. Match with the options The value 1.24×10−3 nm1.24 \times 10^{-3}\ \text{nm} is extremely close to 1.2×10−3 nm1.2 \times 10^{-3}\ \text{nm} — the slight difference is due to rounding hchc 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=hfE = hf and then forget that c=fλc = f\lambda, or to mix up units (e.g., using hc=1240 eV⋅nmhc = 1240\ \text{eV·nm} but then treating the energy in MeV without converting to eV). Always convert MeV to eV first: 1 MeV=106 eV1\ \text{MeV} = 10^6\ \text{eV}.

Tip

Memorising hc=1240 eV⋅nmhc = 1240\ \text{eV·nm} saves enormous time. For any photon energy in eV, the wavelength in nm is simply 1240/E1240/E. For MeV energies, just shift the decimal: 1240/106=1.24×10−31240 / 10^6 = 1.24 \times 10^{-3}.

✓Final answer

The correct option is (B) 1.2×10−3 nm1.2 \times 10^{-3}\ \text{nm}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←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…