Threshold wavelength of a photometal is 24800 Aº. If radiations of energies 1.5eV and 2eV are
incident on it. Ratio between maximum speeds of ejected electrons is
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Answer:
The ratio of the maximum velocity of the ejected electrons is .
Explanation:
We will use the photoelectric equation to solve this question,
(1)
Ф=energy of incident radiation
Ф₀=work function
Kmax=maximum kinetic energy=1\2×m×V²max
First, we need to find the work function of the photo metal in eV i.e.
(2)
The values given in the question are,
λ₀=24800 Aº
Ф₁=1.5eV
Ф₂=2eV
Now,
(3)
We can also write equation (1) as,
(4)
Now as per the question,
(5)
By substituting the required values in equation (5) we get;
Hence, the ratio of the maximum velocity of the ejected electrons is .
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