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Photons of energy 1eV and 2.5 eV successively illuminate a metal whose work function is 0. 5 eV. The ratio of the maximum speeds of the electrons emitted will be
- a)1 : 4
- b)1 : 2
- c)2 : 1
- d)4 : 1
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Photons of energy 1eV and 2.5 eV successively illuminate a metal whose...
Work function=0.5
for 1ev photon
1ev=0.5+kinetic energy
k.E=0.5ev
for 2.5ev photon
2.5ev=0.5ev+k.E
k.E=2ev
ratio of above2 k.E give ratio of square of velocity =1:4
taking square root
finelly the ratio of velocity =1:2ans
for 1ev photon
1ev=0.5+kinetic energy
k.E=0.5ev
for 2.5ev photon
2.5ev=0.5ev+k.E
k.E=2ev
ratio of above2 k.E give ratio of square of velocity =1:4
taking square root
finelly the ratio of velocity =1:2ans
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Photons of energy 1eV and 2.5 eV successively illuminate a metal whose...
Solution:
Given:
Energy of photon 1= 1 eV
Energy of photon 2= 2.5 eV
Work function (ϕ)= 0.5 eV
To find: Ratio of maximum speeds of electrons emitted
Formula used:
The maximum kinetic energy of the photoelectrons is given by
K.E. = hν - ϕ
where h is Planck's constant, ν is the frequency of the incident photon, and ϕ is the work function of the metal.
The maximum speed of the photoelectrons is given by
vmax = √2K.E./m
where m is the mass of the electron.
Calculation:
Using the formula K.E. = hν - ϕ
For photon 1, K.E. = (6.63 x 10^-34 Js x 1.6 x 10^-19 C) x (1 x 10^6 eV / 1 J) x (1/1) - 0.5 eV = 0.5 eV
For photon 2, K.E. = (6.63 x 10^-34 Js x 1.6 x 10^-19 C) x (2.5 x 10^6 eV / 1 J) x (1/1) - 0.5 eV = 2 eV
Using the formula vmax = √2K.E./m
For photon 1, vmax = √(2 x 0.5 eV x 1.6 x 10^-19 J/eV) / 9.1 x 10^-31 kg = 1.04 x 10^6 m/s
For photon 2, vmax = √(2 x 2 eV x 1.6 x 10^-19 J/eV) / 9.1 x 10^-31 kg = 1.64 x 10^6 m/s
Ratio of maximum speeds of electrons emitted = vmax(2.5 eV photon) / vmax(1 eV photon) = 1.64 x 10^6 m/s / 1.04 x 10^6 m/s = 1.58 ≈ 1.6
Therefore, the ratio of maximum speeds of the electrons emitted is 1 : 1.6, which is approximately equal to 1 : 2.
Therefore, option B is the correct answer.
Given:
Energy of photon 1= 1 eV
Energy of photon 2= 2.5 eV
Work function (ϕ)= 0.5 eV
To find: Ratio of maximum speeds of electrons emitted
Formula used:
The maximum kinetic energy of the photoelectrons is given by
K.E. = hν - ϕ
where h is Planck's constant, ν is the frequency of the incident photon, and ϕ is the work function of the metal.
The maximum speed of the photoelectrons is given by
vmax = √2K.E./m
where m is the mass of the electron.
Calculation:
Using the formula K.E. = hν - ϕ
For photon 1, K.E. = (6.63 x 10^-34 Js x 1.6 x 10^-19 C) x (1 x 10^6 eV / 1 J) x (1/1) - 0.5 eV = 0.5 eV
For photon 2, K.E. = (6.63 x 10^-34 Js x 1.6 x 10^-19 C) x (2.5 x 10^6 eV / 1 J) x (1/1) - 0.5 eV = 2 eV
Using the formula vmax = √2K.E./m
For photon 1, vmax = √(2 x 0.5 eV x 1.6 x 10^-19 J/eV) / 9.1 x 10^-31 kg = 1.04 x 10^6 m/s
For photon 2, vmax = √(2 x 2 eV x 1.6 x 10^-19 J/eV) / 9.1 x 10^-31 kg = 1.64 x 10^6 m/s
Ratio of maximum speeds of electrons emitted = vmax(2.5 eV photon) / vmax(1 eV photon) = 1.64 x 10^6 m/s / 1.04 x 10^6 m/s = 1.58 ≈ 1.6
Therefore, the ratio of maximum speeds of the electrons emitted is 1 : 1.6, which is approximately equal to 1 : 2.
Therefore, option B is the correct answer.
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Photons of energy 1eV and 2.5 eV successively illuminate a metal whose work function is 0. 5 eV. The ratio of the maximum speeds of the electrons emitted will bea)1 : 4b)1 : 2c)2 : 1d)4 : 1Correct answer is option 'B'. Can you explain this answer?
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