Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Graph between stopping potential and frequency of light as shown in figure. The minimum energy for ejection of electrons from its surface is ....

solution : Threshold frequency , f₀ = 5 × 10¹⁴ Hz

so, Threshold energy , Φ = hf₀

= 6.62 × 10¯³⁴ Js × 5 × 10¹⁴ Hz

= 33.1 × 10¯²⁰ J

= 3.31 × 10^-19 J

we know, 1eV = 1.6 × 10^-19 J

so, Φ = (3.31 × 10^-19)/(1.6 × 10^-19)

= 2.06 eV

we know, the minimum energy required for ejection of electrons from metal's surface is known as Threshold energy or work function.

therefore the minimum energy for ejection from its surface is 2.06 eV . i.e., option (4) 2.04 eV (it is nearest value of 2.06 eV)