JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Modern Physics

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy

T_{A} eV

and de-Broglie wavelength

λ_{A}

. The maximum kinetic energy of photoelectrons liberated form another metal B by photon of energy 4.50 eV is

T_{B} = (T_{A} - 1.5) eV

. If the de-Broglie wavelength of these photoelectrons

λ_{B} = 2 λ_{A}

, then the work function of metal

B

Answer: (A)

Solution

The de-Broglie wavelength of a photoelectron is

λ = h / 2 m_{e} (K E)

, so

λ \propto 1/ K E

. With

λ_{B} = 2 λ_{A}

\frac{λ _{A}}{λ _{B}} = \frac{K E _{B}}{K E _{A}} = \frac{1}{2}

, and since

K E_{B} = T_{A} - 1.5

this becomes

\frac{1}{2} = \frac{T _{A} - 1.5}{T _{A}}

. Squaring gives

\frac{1}{4} = \frac{T _{A} - 1.5}{T _{A}}

, so

T_{A} = 4 (T_{A} - 1.5)

, which yields

T_{A} = 2

eV and hence

K E_{B} = 2 - 1.5 = 0.5

eV. The photoelectric equation for metal

B

, struck by a

4.5

eV photon, then gives the work function

ϕ_{B} = 4.5 - 0.5 = 4

eV.

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option