Advertisements
Advertisements
प्रश्न
A 3310 Å photon liberates an electron from a material with energy 3 × 10−19 J while another 5000 Å photon ejects an electron with energy 0.972 × 10−19 J from the same material. Determine the value of Planck’s constant and the threshold wavelength of the material.
उत्तर
The energy of ejected electron is given by E = `"hc"/λ - "hc"/λ_0`
`3 xx 10^-19 = "hc" [1/(3310 xx 10^-10) - 1/λ_0]` .......(1)
`0.972 xx 10^-19 = "hc" [1/(5000 xx 10^-10) - 1/λ_0]` .......(2)
Subtracting (2) from (1), we get
`(3 - 0.972) xx 10^-19 = "hc"/10^-10 [1/3310 - 1/5000]`
`2.028 xx 10^-19 = ("h" xx 3 xx 10^8)/10^-10 [1690/(3310 xx 5000)]`
h = `(2.028 xx 10^-19 xx 10^-19 xx 3310 xx 5000)/(3 xx 10^8 xx 1690)`
h = 6.62 × 10−34 Js
Now W0 = `"hc"/λ - "E"`
= `((6.62 xx 10^-34 xx 3 xx 10^8)/(3310 xx 10^-10)) - 3 xx 10^-19`
= (6 − 3) × 10−19
W0 = 3 × 10−19 J
Threshold Wavelength,
λ0 = `"hc"/"W"_0`
= `(6.62 xx 10^-34 xx 3 xx 10^8)/(3 xx 10^-19)`
= 6.62 × 10−7 m
λ0 = 6620 × 10−10 m
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
A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and `λ/2`. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the material 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 threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is __________.
A light of wavelength 500 nm is incident on a sensitive metal plate of photoelectric work function 1.235 eV. The kinetic energy of the photoelectrons emitted is (Take h = 6.6 × 10–34 Js)
Define stopping potential.
Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.
How many photons of frequency 1014 Hz will make up 19.86 J of energy?
When a light of frequency 9 × 1014 Hz is incident on a metal surface, photoelectrons are emitted with a maximum speed of 8 × 105 ms−1. Determine the threshold frequency of the surface.
When a 6000 Å light falls on the cathode of a photo cell, photoemission takes place. If a potential of 0.8 V is required to stop emission of electron, then determine the
- frequency of the light
- energy of the incident photon
- work function of the cathode material
- threshold frequency and
- net energy of the electron after it leaves the surface.
