Question

The work functions of three metals \(A\), \(B\) and \(C\) are respectively \(2.25\), \(2.42\) and \(3.6\) eV. All the three metals are irradiated with light of wavelength \(330\) nm. The kinetic energies of photoelectrons emitted from \(A\), \(B\) and \(C\) are respectively \(E_A\), \(E_B\) and \(E_C\). The correct relationship among \(E_A\), \(E_B\) and \(E_C\) is \[ (h=6.6\times10^{-34}\ \mathrm{Js},\; c=3\times10^8\ \mathrm{ms^{-1}}) \]

• A. \(E_A>E_B>E_C\)
• B. \(E_A=E_B=E_C\)
• C. \(E_A<E_B<E_C\)
• D. \(E_C>E_A>E_B\)

Hint:

Einstein's photoelectric equation is \[ \boxed{ K_{\max}=h\nu-\phi. } \] For the same incident light, a smaller work function produces a larger maximum kinetic energy.

Answer 1

The Correct Option is A

Step 1: Calculate the energy of the incident photon. The photon energy is \[ E=\frac{hc}{\lambda}. \] Using \[ E=\frac{1240}{330}\text{ eV}, \] we get \[ E\approx3.76\text{ eV}. \]

Step 2: Apply Einstein's photoelectric equation. The maximum kinetic energy is \[ K_{\max}=E-\phi. \] For metal \(A\), \[ E_A=3.76-2.25=1.51\text{ eV}. \] For metal \(B\), \[ E_B=3.76-2.42=1.34\text{ eV}. \] For metal \(C\), \[ E_C=3.76-3.60=0.16\text{ eV}. \]

Step 3: Compare the kinetic energies. Clearly, \[ E_A>E_B>E_C. \] Hence, \[ \boxed{E_A>E_B>E_C.} \] Thus, \[ \boxed{(A)} \] is the correct answer.