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PhysicsModern PhysicsPhotoelectric effectHard2 minQB

PhysicsHardnumerical

Passage / Comprehension

When a surface is irradiated with light of wavelength $4950 \overset{˚}{A}$ , a photocurrent appears which vanishes if a retarding potential greater than 0.6 volts is applied across the phototube. When a second source of light is used, it is found that the retarding potential is changed to 1.1 volt. $(hc = 12375 eV - \overset{˚}{A})$

The wavelength (in Å) of the second source is:

Answer:

4125.00

Solution:

Source 1: $\frac{12375}{4950} - ϕ = 0.6$ , so $ϕ = 1.9 e V$

Source 2: $\frac{12375}{λ} - 1.9 = 1.1$ , so $λ = \frac{12375}{3.0} = 4125 \overset{˚}{A}$

Answer 4125.00

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: QB

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