JEE Challenger
More from Dual Nature of Radiation and Matter
PhyHardAdv 2009 • Paper 2

Photoelectric Effect Characteristic Curves for Multi-Wavelength Illumination

Photoelectric effect experiments are performed using three different metal plates p\text{p}, q\text{q} and r\text{r} having work functions ϕp=2.0 eV\phi_p = 2.0\text{ eV}, ϕq=2.5 eV\phi_q = 2.5\text{ eV} and ϕr=3.0 eV\phi_r = 3.0\text{ eV}, respectively. A light beam containing wavelengths of 550 nm550\text{ nm}, 450 nm450\text{ nm} and 350 nm350\text{ nm} with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is
[Take hc=1240 eV nmhc = 1240\text{ eV nm}]

Options

A
Option A Diagram 1
Correct
B
Option B Diagram 1
C
Option C Diagram 1
D
Option D Diagram 1

Topics & Concepts

Dual Nature of Radiation and MatterPhotoelectric Effect

Step-by-Step Solution

To determine the correct II-VV characteristic curves for the three plates, we first find the energy of the incident photons of each wavelength using E=hcλE = \frac{hc}{\lambda}, with hc=1240 eV nmhc = 1240\text{ eV nm}:

  1. For λ1=550 nm\lambda_1 = 550\text{ nm}:
    E1=1240550 eV≈2.25 eVE_1 = \frac{1240}{550}\text{ eV} \approx 2.25\text{ eV}

  2. For λ2=450 nm\lambda_2 = 450\text{ nm}:
    E2=1240450 eV≈2.76 eVE_2 = \frac{1240}{450}\text{ eV} \approx 2.76\text{ eV}

  3. For λ3=350 nm\lambda_3 = 350\text{ nm}:
    E3=1240350 eV≈3.54 eVE_3 = \frac{1240}{350}\text{ eV} \approx 3.54\text{ eV}

The work functions of the three metal plates are:

  • ϕp=2.0 eV\phi_p = 2.0\text{ eV}
  • ϕq=2.5 eV\phi_q = 2.5\text{ eV}
  • ϕr=3.0 eV\phi_r = 3.0\text{ eV}

Condition for Photoelectric Emission:

Photoelectric emission occurs only when the incident photon energy is greater than the work function of the metal (E>ϕE > \phi):

  • For plate p\text{p} (ϕp=2.0 eV\phi_p = 2.0\text{ eV}):

    • E1=2.25 eV>2.0 eVE_1 = 2.25\text{ eV} > 2.0\text{ eV} (causes emission)
    • E2=2.76 eV>2.0 eVE_2 = 2.76\text{ eV} > 2.0\text{ eV} (causes emission)
    • E3=3.54 eV>2.0 eVE_3 = 3.54\text{ eV} > 2.0\text{ eV} (causes emission)
      All three wavelengths contribute to the photocurrent.

  • For plate q\text{q} (ϕq=2.5 eV\phi_q = 2.5\text{ eV}):

    • E1=2.25 eV<2.5 eVE_1 = 2.25\text{ eV} < 2.5\text{ eV} (no emission)
    • E2=2.76 eV>2.5 eVE_2 = 2.76\text{ eV} > 2.5\text{ eV} (causes emission)
    • E3=3.54 eV>2.5 eVE_3 = 3.54\text{ eV} > 2.5\text{ eV} (causes emission)
      Two wavelengths (λ2\lambda_2 and λ3\lambda_3) contribute to the photocurrent.

  • For plate r\text{r} (ϕr=3.0 eV\phi_r = 3.0\text{ eV}):

    • E1=2.25 eV<3.0 eVE_1 = 2.25\text{ eV} < 3.0\text{ eV} (no emission)
    • E2=2.76 eV<3.0 eVE_2 = 2.76\text{ eV} < 3.0\text{ eV} (no emission)
    • E3=3.54 eV>3.0 eVE_3 = 3.54\text{ eV} > 3.0\text{ eV} (causes emission)
      Only one wavelength (λ3\lambda_3) contributes to the photocurrent.

Saturation Current (IsatI_{\text{sat}}):

The saturation current is proportional to the number of photoelectrons emitted per second. Since each wavelength beam has equal intensity:
Isat,p>Isat,q>Isat,rI_{\text{sat}, p} > I_{\text{sat}, q} > I_{\text{sat}, r}

Stopping Potential (V0V_0):

The stopping potential is determined by the maximum kinetic energy of the emitted photoelectrons, which is produced by the highest energy photon (E3=3.54 eVE_3 = 3.54\text{ eV}):
eV0=Kmax⁡=E3−ϕe V_0 = K_{\max} = E_3 - \phi

  • For plate p\text{p}:
    eV0,p=3.54 eV−2.0 eV=1.54 eV  ⟹  V0,p=1.54 Ve V_{0, p} = 3.54\text{ eV} - 2.0\text{ eV} = 1.54\text{ eV} \implies V_{0, p} = 1.54\text{ V}
  • For plate q\text{q}:
    eV0,q=3.54 eV−2.5 eV=1.04 eV  ⟹  V0,q=1.04 Ve V_{0, q} = 3.54\text{ eV} - 2.5\text{ eV} = 1.04\text{ eV} \implies V_{0, q} = 1.04\text{ V}
  • For plate r\text{r}:
    eV0,r=3.54 eV−3.0 eV=0.54 eV  ⟹  V0,r=0.54 Ve V_{0, r} = 3.54\text{ eV} - 3.0\text{ eV} = 0.54\text{ eV} \implies V_{0, r} = 0.54\text{ V}

Thus, in terms of stopping potential magnitude:
V0,p>V0,q>V0,rV_{0, p} > V_{0, q} > V_{0, r}
On the negative potential axis, the current becomes zero at −V0-V_0. Therefore, plate p\text{p} cuts off furthest to the left, followed by q\text{q}, and plate r\text{r} cuts off closest to the origin.

Combining both results:

  • Stopping potential: ∣−V0,p∣>∣−V0,q∣>∣−V0,r∣|-V_{0, p}| > |-V_{0, q}| > |-V_{0, r}|
  • Saturation current: Ip>Iq>IrI_p > I_q > I_r

This corresponds to graph (A).

PreviousNext