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Atoms and Nuclei question

2022 · 29 Jul · Shift 1 · Q64
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Atoms and Nuclei question

2022 · 29 Jul · Shift 1 · Q64

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
  1. A
    3 : 4
  2. B
    4 : 3
  3. C
    1 : 4
  4. D
    4 : 1
View written solutionFree

Correct answer: A

  1. Energy levels of hydrogen atom

    For hydrogen, the energy of the electron in the nthn^{\text{th}}nth orbit is En=−13.6n2 eVE_n = -\frac{13.6}{n^2}\,\text{eV}En​=−n213.6​eV

  2. Photon energy for transition from second level to first level

    Here, the transition is n=2→n=1n=2 \to n=1n=2→n=1.

    E2=−13.64=−3.4 eVE_2 = -\frac{13.6}{4} = -3.4\,\text{eV}E2​=−413.6​=−3.4eV E1=−13.6 eVE_1 = -13.6\,\text{eV}E1​=−13.6eV

    Energy of emitted photon: hν1=E2−E1h\nu_1 = E_2 - E_1hν1​=E2​−E1​ hν1=(−3.4)−(−13.6)=10.2 eVh\nu_1 = (-3.4) - (-13.6) = 10.2\,\text{eV}hν1​=(−3.4)−(−13.6)=10.2eV

  3. Photon energy for transition from highest permitted level to first level

    The highest permitted level means n=∞n=\inftyn=∞.

    E∞=0E_{\infty} = 0E∞​=0

    Transition is n=∞→n=1n=\infty \to n=1n=∞→n=1.

    Energy of emitted photon: hν2=E∞−E1h\nu_2 = E_{\infty} - E_1hν2​=E∞​−E1​ hν2=0−(−13.6)=13.6 eVh\nu_2 = 0 - (-13.6) = 13.6\,\text{eV}hν2​=0−(−13.6)=13.6eV

  4. Required ratio

    hν1:hν2=10.2:13.6h\nu_1 : h\nu_2 = 10.2 : 13.6hν1​:hν2​=10.2:13.6

    Divide by 3.43.43.4: =3:4= 3 : 4=3:4

  5. Option check

    • A: 3:43:43:4 ✅
    • B: 4:34:34:3 ❌
    • C: 1:41:41:4 ❌
    • D: 4:14:14:1 ❌

Therefore, the correct answer is A.

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