Advertisements
Advertisements
प्रश्न
Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.
Advertisements
उत्तर
V0 = 81 V
e = 1.6 × 10−19 C
m = 9.1 × 10−31 kg
Maximum kinetic energy of electron,
Kmax = `"eV"_0`
= 1.6 × 10−19 × 81
= 129.6 × 10−19
= 1.29 × 10−17
Kmax = 1.3 × 10−17 J
Maximum velocity of photoelectron,
vmax = `sqrt((2"eV"_0)/"m")`
= `sqrt((2 xx 1.6 xx 10^-19 xx 81)/(9.1 xx 10^-31))`
= `sqrt((259.2 xx 10^-19)/(9.1 xx 10^-31))`
= `sqrt(28.48 xx 10^12)`
vmax = 5.3 × 106 ms−1
APPEARS IN
संबंधित प्रश्न
When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is `"V"/4`. The threshold wavelength for the metallic surface is
In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be
The work functions for metals A, B and C are 1.92 eV, 2.0 eV and 5.0 eV respectively. The metal/metals which will emit photoelectrons for a radiation of wavelength 4100Å is/are
Mention the two features of x-ray spectra, not explained by classical electromagnetic theory.
Explain the effect of potential difference on photoelectric current.
List out the laws of photoelectric effect.
Explain the quantum concept of light.
Give the construction and working of photo emissive cell.
At the given point of time, the earth receives energy from the sun at 4 cal cm–2 min–1. Determine the number of photons received on the surface of the Earth per cm2 per minute. (Given: Mean wavelength of sunlight = 5500 Å)
UV light of wavelength 1800 Å is incident on a lithium surface whose threshold wavelength is 4965 Å. Determine the maximum energy of the electron emitted.
