Photons of frequency f is incident on a metal surface of threshold frequency f0 . The maximum K.E. of emitted photoelectrons is
The photoelectric threshold frequency of a metal is f0. When light of frequency 4f0is incident on the metal, the maximum K.E. of the emitted electron is
The maximum kinetic energy of the emitted electrons is given by
Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential is
λ for U.V is less than λ for visible light
ν for U.V is greater than ν for visible light
∴ potential is greater for U.V light.
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material .If the frequency is halved and the intensity is doubled, the photoelectric current becomes
When the frequency of incident light is halved of its original value i.e. 1.5v0 then it becomes less than the threshold value. In that case no photoelectric effect takes place. No photoelectrons would be emitted. The photoelectric current becomes zero.
Light of wavelength 4000 Â is incident on a metal of work function 3.2 x 10-19 J. What is the maximum kinetic energy of the emitted electron?
λ= 4000 Å ;W•= 3.2× 10-19 j
as we know that
K.E = (hc/ λ) - W•
= (6.6× 10-34 × 3 × 10-19/ 4000×10-10 )- 3.20× 10-19
= 1.75 × 10-19
Light from a bulb is falling on a wooden table but no photo electrons are emitted as
It is because the work function of wood is more than the energy of the incident photons of light. Hence, photoelectrons are not emitted from the wooden table.
Photons of energy 6eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?
From photo-electric equation, eV0= E−φ
V0= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV. What is the stopping potential?
Stopping potential (Vo) is given by
Vo=W/q where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Light of two different frequencies whose photons have energies of 1eV and 2.5 eV respectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons is
For maximum speed of the photo electrons (½)mv2=Ephoton−ϕ
where ϕ=0.5 eV is the work function of the metal.
Energy of photon E1=1eV
We get (½) mv12=0.5 eV ....(1)
Energy of photon E2=2.5eV
∴ (½) mv22=2.5−0.5
We get (½)mv22=2 eV ....(2)
Ratio of maximum speeds v12/ v22=0.5/2=1/4
We get v1/v2 =1/2
A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the:
for photoelectric effect,
hf=hf0+Ek, Here f>f0