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प्रश्न
If the total energy of radiation of frequency 1014 Hz is 6.63 J, Calculate the number of photons in the radiation.
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उत्तर
Given:
E = 6.63 J, ν = 1014 Hz,
We know, h = 6.63 × 10–34 Js.
To find: Number of photons (n)
Formula: n = `"E"/"hv"`
Calculation:
Using formula,
n = `6.63/(6.63 xx 10^-34 xx 10^14)`
∴ n = 1020
The number of photons emitted in the radiation is 1020.
संबंधित प्रश्न
Choose the correct option.
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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 |
The threshold wavelength of tungsten is 2.76 x 10-5 cm.
(a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 x 10-5 cm.
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(i) if ultraviolet radiation of wavelength λ = 1.80 × 10-5 cm and
(ii) radiation of frequency 4 x 1015 Hz is made incident on the tungsten surface?
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| Incident wavelength (in Å) | 2536 | 3650 |
| Stopping potential (in V) |
1.95 | 0.5 |
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[Energy of electron in nth orbit = `-13.6/"n"^2` eV ]
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