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JEE Advanced Physics Dual Nature of Matter 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Physics Question (2023) — Solution

Question

A Hydrogen-like atom has atomic number Z . Photons emitted in the electronic transitions from level n = 4 to level n = 3 in these atoms are used to perform photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is 1 . 95   eV . If the photoelectric threshold wavelength for the target metal is 310   nm , the value of Z is _____. [Given h c = 1240   eV - nm and R h c = 13 . 6   eV , where R is the Rydberg constant, h is the Planck’s constant and c  is the speed of light in vacuum]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

From Einstein's photoelectric effect, we can write E = K E m a x + ϕ ⇒ E = K E m a x + h c λ ⇒ ∆ E 4   to   3 = 1 . 95   eV + 1240 310 eV ⇒ ∆ E 4   to   3 = 5 . 95   eV Now, as we know for Hydrogen like atoms, we can write  ∆ E = 13 . 6   Z 2 1 n 1 2 - 1 n 2 2 Therefore, ⇒ 13 . 6   Z 2 1 3 2 - 1 4 2 = 5 . 95 ⇒ 13 . 6   Z 2 7 9 × 16 = 5 . 95 ⇒ Z 2 = 5 . 95 × 9 × 16 13 . 6 × 7 = 9 ⇒ Z = 3

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