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प्रश्न
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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उत्तर
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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संबंधित प्रश्न
Using the values of work function given in the following table, tell which metal will require the highest frequency of incident radiation to generate photocurrent.
Typical values of work function for some common metals
| Metal | Work function (in eV) |
| Potassium | 2.3 |
| Sodium | 2.4 |
| Calcium | 2.9 |
| Zinc | 3.6 |
| Silver | 4.3 |
| Aluminium | 4.3 |
| Tungsten | 4.5 |
| Copper | 4.7 |
| Nickel | 5.0 |
| Gold | 5.1 |
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