Physics

Problem

When photon of energy

4.0 \text{ eV}

strikes the surface of a metal

A

, the ejected photoelectrons have maximum kinetic energy

T_A \text{ eV}

and de-Broglie wavelength

\lambda_A

. The maximum kinetic energy of photoelectrons liberated from another metal

B

by photon of energy

4.50 \text{ eV}

T_B = (T_A - 1.5) \text{ eV}

. If the de-Broglie wavelength of these photoelectrons

\lambda_B = 2\lambda_A

, then the work function of metal

B

is: (a)

4 \text{ eV}

(b)

2 \text{ eV}

(c)

1.5 \text{ eV}

(d)

3 \text{ eV}