In a photoelectric experiment, with light of wavelength `lambda`, the fastest election has speed v. If the exciting wavelength is changed to `(3lambda)/4`, the speed of the fastest emitted electron will become
A
`= v((4)/(3))^(1/2)`
B
`= v((3)/(4))^(1/2)`
C
`gt v((4)/(3))^(1/2)`
D
`lt v((4)/(3))^(1/2)`
Text Solution
Verified by Experts
The correct Answer is:
C
`hv_(0)^(2) - hv_(0) = (1)/(2) m v^(2)` `:. (4)/(3) hv_(0) - hv_(0) = (1)/(2) m v^(-2)` `:. (v^(2))/(v^(2) = ((4)/(3) v - v_(0)))/(v - v_(0)) :. V ^(1) = v sqer(((4)/(3) v - v_(0)))/(v - v_(0)))` `:. V^(1) gt sqrt((4)/(3))`
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise Comprehension Based Questions|2 Videos
MOVING CHARGES AND MAGNETISM
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise MCQs(d )|1 Videos
Similar Questions
Explore conceptually related problems
In a photoelectric experiment, with light of wavelength lamda , the fastest electron has speed v. If the exciting wavelength is changed to 5lamda/4 , the speed of the fastest emitted electron will become
In a photoemissive cell, with exciting wavelength lambda , the faster electron has speed v. If the exciting wavelength is changed to 3lambda//4 , the speed of the fastest electron will be
In a photomissive cell with executing wavelength lamda//2 , work function phi , the fastest moving electron has acquired speed as v. If the executing wavelength is changed to (5lamda)/4 , the speed of the fastest emitted electron will be
In a photo-emissive cell, with exciting wavelength lambda , the maximum kinetic energy of electron is K. If the exciting wavelength is changed to ( 3 lambda)/( 4) the kinetic energy of the fastest emitted electron will be :
In a photo-emissive cell, with exciting wavelength lambda_1 , the maximum kinetic energy of electron is K. If the exciting wavelength is changed to lambda_2 , the maximum kinetic energy of electron is 2K then
In photoelectric effect experiment, the stopping potential for incident light of wavelength 4000 Å, is 3V. If the wavelength is changed to 2500 Å, the stopping potential will be
In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be ________eV Given E (in eV) = (1237)/(lambda ("in mm"))
If a hydrogen atom at rest, emits a photon of wavelength lambda , the recoil speed of the atom of mass m is given by :
For a photoelectric experiment graph between kinetic energy of fastest electron and frequency of light is plotted find threshold wavelength of metal.
SUNIL BATRA (41 YEARS IITJEE PHYSICS)-MODERN PHYSICS-MCQ (One Correct Answer
