Test: Photoelectric Effect - JEE MCQ

K_max=eV₀=hf-hf₀
K_max=eV₀= h(f-f₀)

The maximum kinetic energy of the emitted electrons is given by
K_max​=hυ−ϕ₀​=h(4υ)−h(υ)=3hυ

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.
as K.Eα1​/λ

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.

Given,
λ= 4000 Å ;W=3.2× 10^-19 j
as we know that
K.E = (hc/ λ) - W
= (6.6× 10^-34 × 3 × 10⁸/ 4000×10^-10 )- 3.20× 10^-19 
= 1.75 × 10^-19
 

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.

From photo-electric equation,   eV₀​= E−φ
 eV₀​=(6−2.1)eV
V_0​= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
 

Stopping potential (V_o​) is given by
V_o​=W/q​ where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Hence, V_o​=(10eV)/e​=10V

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
∴  (½)​mv12​=1−0.5
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

By theory
for photoelectric effect,
hf=hf₀+E_k, Here f>f₀