A light of wavelength 3540 mathringA falls on a metal having work function of 2.5 eV. The ejected electron collides with another target metal inelastically and its total kinetic energy is utilized to raise the temperature of target metal. The mass of target metal is 10-3 kg and its specific heat is 160 J kg -1 ° C -1. If 1018 electrons are ejected per second, then find the rate of raise of temperature (. in . ° C / s ) of the metal. (Assume there is no loss of energy of ejected electron by any other process, all the electron are reaching the target metal with maximum kinetic energy and h c=12400 eV mathringA )

Question

A light of wavelength 3540 mathringA falls on a metal having work function of 2.5 eV. The ejected electron collides with another target metal inelastically and its total kinetic energy is utilized to raise the temperature of target metal. The mass of target metal is 10-3 kg and its specific heat is 160 J kg -1 ° C -1. If 1018 electrons are ejected per second, then find the rate of raise of temperature (. in . ° C / s ) of the metal. (Assume there is no loss of energy of ejected electron by any other process, all the electron are reaching the target metal with maximum kinetic energy and h c=12400 eV mathringA ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Energy of photon =(h c/λ)=(12400/3540) eV ≈ 3.5 eV Thus, K E max of ejected electron =1 eV Thus, energy lost per electron =1 eV =1 × 1.6 × 10-19 J Total energy gained by target metal per second =1.6 × 10-19 × 1018 J =0.16 J Thus, 0.16=m s (d T/d t) ⇒ ⋅ (d T/d t)=(0.16/10-3 × 160)° C / s

Energy of photon =λhc​=354012400​eV≈3.5eV Thus, KEmax​ of ejected electron =1eV Thus, energy lost per electron =1eV=1×1.6×10−19J Total energy gained by target metal per second =1.6×10−19×1018J=0.16J Thus, 0.16=msdtdT​ ⇒⋅dtdT​=10−3×1600.16​∘C/s