Question

Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the materials is(a) 4 × 10¹⁵ Hz(b) 5 × 10¹⁵ Hz(c) 1.6 × 10¹⁵ Hz(d) 2.5 × 10¹⁵ Hz

Answer 1

answer : option (c) 1.6 × 10¹⁵ Hz

explanation : energy released from emission of electron, E = -3.4eV -(-13.6eV) = 10.2eV

from photoelectric equation,

$K.E_{max}=E-\phi$

we know, maximum kinetic energy equals to product of stopping potential and charge of electron.

e.g., $K.E_{max}=eV_0=3.57eV$

so, $\phi=E-K.E_{max}$

= 10.2eV - 3.57eV = 6.63eV

hence, $\phi=6.63eV=6.63\times1.6\times10^{-19}J$

now, Threshold energy = plank's constant × Threshold frequency

so, 6.63 × 1.6 × 10^-19 = 6.63 × 10^-34 × Threshold frequency

hence, Threshold frequency = 1.6 × 10^15 Hz

Answer 2

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♦ Option [B] is Correct

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