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JEE Physics 2024 Question with Solution

Hydrogen atom is bombarded with electrons accelerated through a potential difference VV, which causes excitation of hydrogen atoms. If the experiment is performed at T=0 KT = 0 \, \text{K}, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be α/10 V\alpha/10 \, \text{V}, where α\alpha is:

  • A

    121121

  • B

    150150

  • C

    135135

  • D

    110110

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Hydrogen atoms are initially in the ground state because the experiment is performed at T=0 KT = 0 \, \text{K}. The minimum potential difference is written as α/10 V\alpha/10 \, \text{V}.

Find: The value of α\alpha required to observe any Balmer series line.

To observe a Balmer line, the atom must first be excited to at least n=3n = 3, because Balmer series corresponds to transitions ending at n=2n = 2 and starting from n≥3n \ge 3.

Using the hydrogen energy levels,

En=−13.6n2 eVE_n = -\frac{13.6}{n^2} \, \text{eV}

So,

E1=−13.6 eV,E3=−13.69 eVE_1 = -13.6 \, \text{eV}, \qquad E_3 = -\frac{13.6}{9} \, \text{eV}

The minimum excitation energy needed is from n=1n = 1 to n=3n = 3:

ΔE=E3−E1=−13.69−(−13.6)\Delta E = E_3 - E_1 = -\frac{13.6}{9} - (-13.6) ΔE=13.6(1−19)=13.6×89≈12.09 eV\Delta E = 13.6\left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} \approx 12.09 \, \text{eV}

Hence the minimum accelerating potential is

Vmin⁡=12.09 VV_{\min} = 12.09 \, \text{V}

Given that

Vmin⁡=α10 VV_{\min} = \frac{\alpha}{10} \, \text{V}

therefore,

α10=12.09\frac{\alpha}{10} = 12.09 α≈120.9≈121\alpha \approx 120.9 \approx 121

Therefore, the correct option is A.

The solution concludes α=18.9\alpha = 18.9 by using the transition energy from n=3n = 3 to n=2n = 2, but that is the emitted photon energy, not the minimum excitation energy needed from the ground state at T=0 KT = 0 \, \text{K}. The option-supported and physically correct result is α=121\alpha = 121.

Energy-gap Insight

Given: At T=0 KT = 0 \, \text{K}, all hydrogen atoms are in the ground state n=1n = 1.

Find: The least energy needed so that a Balmer transition can occur.

A Balmer line appears only if an electron reaches at least n=3n = 3 and then falls to n=2n = 2. So do not calculate the emitted photon energy 3→23 \to 2; calculate the excitation energy 1→31 \to 3.

Thus,

ΔE=13.6(1−132)=13.6(89)≈12.1 eV\Delta E = 13.6\left(1 - \frac{1}{3^2}\right) = 13.6\left(\frac{8}{9}\right) \approx 12.1 \, \text{eV}

So,

α10=12.1⇒α≈121\frac{\alpha}{10} = 12.1 \Rightarrow \alpha \approx 121

Hence, the correct option is A.

Common mistakes

  • Using the photon energy of the first Balmer line, 3→23 \to 2, instead of the excitation energy from the ground state. This is wrong because at T=0 KT = 0 \, \text{K} the atoms start in n=1n = 1. You must first excite the atom to at least n=3n = 3.

  • Ignoring the meaning of 'minimum potential difference needed to observe any Balmer series lines'. The question asks for the least incident electron energy required to make a Balmer emission possible, not the energy of the emitted Balmer photon itself.

  • Assuming atoms are already present in excited states. At T=0 KT = 0 \, \text{K}, hydrogen atoms are in the ground state, so the required excitation is 1→31 \to 3, not 2→32 \to 3 or 3→23 \to 2.

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