JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Modern Physics

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals

P, Q

and

R

are

E_{P}, E_{Q}

and

E_{R}

, respectively, and they are related by

E_{P} = 2 E_{Q} = 2 E_{R}

. In this experiment, the same source of monochromatic light is used for metals

P

and

Q

while a different source of monochromatic light is used for the metal

R

. The work functions for metals

P, Q

and

R

are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal

R

, in eV , is ____.

Answer: 6

Solution

Using Einstein's photoelectric equation:

E_{photon} = ϕ + K E_{max}

For metals P and Q (same light source, photon energy

E_{1}

E_{1} = 4.0 + E_{P}

E_{1} = 4.5 + E_{Q}

Given

E_{P} = 2 E_{Q}

4.0 + E_{P} = 4.5 + E_{Q}

4.0 + 2 E_{Q} = 4.5 + E_{Q}

E_{Q} = 0.5 eV

E_{P} = 1.0 eV

Therefore:

E_{1} = 4.0 + 1.0 = 5.0 eV

Given

E_{P} = 2 E_{R}

E_{R} = \frac{E _{P}}{2} = 0.5 eV

For metal R (different light source, photon energy

E_{2}

E_{2} = 5.5 + E_{R} = 5.5 + 0.5 = 6.0 eV

Therefore, the energy of the incident photon for metal R is 6 eV.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Integer answer