The surface of a metal is illuminated alternately with photons of energies E? = 4eV and E? = 2.5eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is
The surface of a metal is illuminated alternately with photons of energies E? = 4eV and E? = 2.5eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is
E? = φ + K?
E? = φ + K?
E? - E? = K? - K?
Now V? /V? = 2
K? /K? = 4; K? = 4K?
Now from equation (2)
⇒ 4 - 2.5 = 4K? - K?
1.5 = 3K?
K? = 0.5eV
Now putting this
Value in equation (2)
2.5 = φ + 0.5eV
φ = 2eV
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