Question

Values of work function (Wo) for a few metals are given below

Metal	Li	Na	K	Mg	Cu	Ag
Wo/eV	2.42	2.3	2.25	3.7	4.8	4.3

The number of metals which will show photoelectric effect when light of wavelength 400nm falls on it is____ 

Given: h = 6.6 × 10^-34 J s 
c = 3 x 10⁸ m s^-1 
e = 1.6 x 10^-19 C

Answer 1

Correct Answer: 3

To determine the number of metals that will exhibit the photoelectric effect when exposed to light of wavelength 400 nm, we must compare the energy of the incident photons to the work function (\(W_o\)) of each metal.

The energy of a photon (\(E\)) is given by:

\(E = \frac{hc}{\lambda}\)

Substituting the given values: \(h = 6.6 \times 10^{-34} \text{ J s}\), \(c = 3 \times 10^8 \text{ m/s}\), and \(\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}\), the energy becomes:

\(E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}}\)

\(E = \frac{19.8 \times 10^{-26}}{400 \times 10^{-9}}\)

\(E = 4.95 \times 10^{-19} \text{ J}\)

Convert this energy to electron volts (eV), using \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\):

\(E = \frac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}} \text{ eV}\)

\(E \approx 3.094 \text{ eV}\)

Next, compare this energy to the work functions of the metals:

• Li: 2.42 eV
• Na: 2.3 eV
• K: 2.25 eV
• Mg: 3.7 eV
• Cu: 4.8 eV
• Ag: 4.3 eV

The photoelectric effect occurs if \(E\) is greater than \(W_o\). Thus, for Li, Na, and K, the condition is satisfied since 3.094 eV > 2.42 eV, 2.3 eV, and 2.25 eV, respectively. However, Mg, Cu, and Ag do not satisfy this condition. Therefore, 3 metals exhibit the photoelectric effect.

The calculated value, 3, fits the given range of 3,3.