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प्रश्न
State Einstein photoelectric equation.
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उत्तर
K.E.max = `(hν - phi_0)`
or K.E.max = (hν – hv0)
or `1/2 mv_(max)^2 = hv - hv_0`
where, (K.E.)max → Maximum kinetic energy
= `1/2 mv_(max)^2`
`phi_0` → Photoelectric work function = hv0
h → Planck’s constant
v → Frequency of radiation
v0 → Threshold frequency
संबंधित प्रश्न
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| Incident wavelength (in Å) | 2536 | 3650 |
| Stopping potential (in V) |
1.95 | 0.5 |
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