Question

# The work function of a metal is 2.5 eV. The maximum kinetic energy of the photoelectrons emitted if a radiation of wavelength 3000Ao falls on it is :(h=6.63×10−34 Js and c=3×108 m/s)

A
1.12×1019J
B
4.8×1019J
C
3.2×1019J
D
2.63×1019J
Solution
Verified by Toppr

#### Maximum kinetic energy, K.E.max=hcλ−ϕ=6.63×10−34×3×1083000×10−10−2.5×1.6×10−19=(6.63×10−19−4×10−19)J=2.63×10−19J.So, the answer is option (D).

3
Similar Questions
Q1

A photon of wavelength 4 × 107 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
(i) the energy of the photon (eV),

(ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV= 1.6020 × 1019 J).

View Solution
Q2

If the work function for a certain metal is 3.2×1019 joule and it is illuminated with light of frequency 8×1014 Hz. The maximum kinetic energy of the photo-electrons would be (h=6.63×1034 Js

View Solution
Q3

The work function of a metal is 1.6×1019J. When the metal surface is illuminated by the light of wavelength 6400A , then the maximum kinetic energy of emitted photo-electrons will be (Planck's constant h=6.4×1034 Js)

View Solution
Q4

When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV. the de - Broglie wavelength of emitted electrons is A. (Round off to the Nearest Interger).

[Use : 3=1.73,h=6.63×1034Js
me=9.1×1031kg;c=3.0×108ms1;1eV=1.6×1019J]

View Solution
Q5

If the work function for a certain metal is 3.2×1019J and it is illuminated with light of frequency 8×1014 Hz . The maximum kinetic energy of the photoelectrons would be: (h=6.63×1034 Js)

View Solution
Solve
Guides