In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function φ=2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. ( h =6.63 × 10-34 J., c =3 × 108 ms -1 )

Question

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function φ=2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. ( h =6.63 × 10-34 J., c =3 × 108 ms -1 ) (A) 1.3 V (B) 1.1 V (C) 1.9 V (D) 0.6 V

Tardigrade Admin · Accepted Answer

KE max = eV S =( hc /λ)-φ ⇒ eV s =(1240/280)-2.5=1.93 eV arrow V S 1=1.93 V... (i) arrow eV S 2=(1240/400)-2.5=0.6 eV ⇒ V S 2=0.6 V ldots (ii) Δ V = V S 1- V S 2 =1.93-0.6=1.33 V

Q. In a photoelectric experiment ultraviolet light of wavelength 280nm is used with lithium cathode having work function ϕ=2.5eV. If the wavelength of incident light is switched to 400nm, find out the change in the stopping potential. (h=6.63×10−34J, c=3×108ms−1 )

1.3 V

1.1 V

1.9 V

0.6 V

KEmax​=eVS​=λhc​−ϕ ⇒eVs​=2801240​−2.5=1.93eV →VS1​​=1.93V... (i) →eVS2​​=4001240​−2.5=0.6eV ⇒VS2​​=0.6V… (ii) ΔV=VS1​​−VS2​​ =1.93−0.6=1.33V

$K E_{m a x} = e V_{S} = \frac{h c}{λ} - ϕ$
$\Rightarrow e V_{s} = \frac{1240}{280} - 2.5 = 1.93 e V$
$\to V_{S_{1}} = 1.93 V$ ... (i)
$\to e V_{S_{2}} = \frac{1240}{400} - 2.5 = 0.6 e V$
$\Rightarrow V_{S_{2}} = 0.6 V \dots$ (ii)
$Δ V = V_{S_{1}} - V_{S_{2}}$
$= 1.93 - 0.6 = 1.33 V$