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Question
The work function of a surface is 3.1 eV. A photon of frequency 1 × 1015 Hz. Is an incident on it. Calculate the incident wavelength is photoelectric emission occurs or not.
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Solution
Given:
φ0 = 3.1 eV = 3.1 × 1.6 × 10−19 J
ν = 1 × 1015 Hz
To find:
- Incident wavelength (λ)
- Will photoelectric emission occur.
Formulae:
- `lambda_0 = "hc"/phi_0`
- `lambda = "c"/"v"`
Calculation:
Using formula (i),
`lambda_0 = (6.63 xx 10^-34 xx 3 xx 10^8)/(3.1 xx 1.6 xx 10^-19)`
= `(6.63 xx 3)/(3.1 xx 1.6) xx 10^-7`
= antilog {log 6.63 + log 3 − log 3.1 − log 1.6} × 10−7
= antilog {0.8215 + 0.4771 − 0.4914 − 0.2041} × 10−7
= antilog {0.6031} × 10−7
= 4.010 × 10−7 m
= 4010 Å
Using formula (ii),
`lambda = (3 xx 10^8)/(1 xx 10^15)`
= 3 × 10−7 m = 3000 Å
As λ < λ0, photoelectric mission will occur.
- Incident wavelength is 3000 Å
- Photoelectric emission will occur.
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