Ultraviolet light of wavelength 350 nm and intensity 1.00 W / m 2 is directed at a potassium surface. What will be the maximum kinetic energy of the photoelectrons? (Work function of potassium =2.2 eV )

Question

Ultraviolet light of wavelength 350 nm and intensity 1.00 W / m 2 is directed at a potassium surface. What will be the maximum kinetic energy of the photoelectrons? (Work function of potassium =2.2 eV ) (A) 3.5 eV (B) 2.6 eV (C) 2.2 eV (D) 1.3 eV

Tardigrade Admin · Accepted Answer

Given, λ=350 nm =350 × 10-7 m Intensity =1.00 ω / m 2 work function W =2.2 eV Energy of the photon =(h c/λ) =(6.6 × 10-34 × 3 × 108/350 × 108) =3.5 eV K max =E-W =(3.5-2.2) eV =1.3 eV

Q. Ultraviolet light of wavelength $350 nm$ and intensity $1.00 W / m^{2}$ is directed at a potassium surface. What will be the maximum kinetic energy of the photoelectrons? (Work function of potassium $= 2.2 e V$ )

3.5 eV

2.6 eV

2.2 eV

1.3 eV

Given, $λ = 350 nm$
$= 350 \times 1 0^{- 7} m$
Intensity $= 1.00 ω / m^{2}$
work function $W = 2.2 e V$
Energy of the photon $= \frac{h c}{λ}$
$= \frac{6.6 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{350 \times 1 0 ^{8}}$
$= 3.5 e V K_{m a x} = E - W$
$= (3.5 - 2.2) e V = 1.3 e V$

Q. Ultraviolet light of wavelength 350nm and intensity 1.00W/m2 is directed at a potassium surface. What will be the maximum kinetic energy of the photoelectrons? (Work function of potassium =2.2eV )