Advertisements
Advertisements
Question
Light of wavelength 2000 Å falls on a metal surface of work function 4.2 eV.
What is the kinetic energy (in eV) of the fastest electrons emitted from the surface?
Advertisements
Solution
λ = 2000 Å = (2000 × 10-10)m
Wo = 4.2eV
h = 6.63 × 10-34 JS
Using Einstein's photoelectric equation
K. E. = (6.2 – 4.2) eV = 2.0 eV
APPEARS IN
RELATED QUESTIONS
An α-particle and a proton are accelerated through the same potential difference. Find the ratio of their de Broglie wavelength.
Calculate the de-Broglie wavelength of an electron moving with one-fifth of the speed of light.Neglect relativistic effects. (`h = 6.63 xx 10^(-34)` J.s, c= `3xx10^8`m/s, mass of electron = `9 xx 10^(-31) kg)`
State de Broglie hypothesis
Show on a graph the variation of the de Broglie wavelength (λ) associated with an electron, with the square root of accelerating potential (V) ?
An electron, an alpha particle and a proton have the same kinetic energy.
Which one of these particles has the largest de-Broglie wavelength?
An electron of energy 150 eV has wavelength of 10-10m. The wavelength of a 0.60 keV electron is?
A proton and an electron are accelerated by the same potential difference. Let λe and λpdenote the de Broglie wavelengths of the electron and the proton, respectively.
Describe in brief what is observed when moving electrons are allowed to fall on a thin graphite film and the emergent beam falls on a fluorescent screen.
For production of characteristic of X-rays the electron trans mission is
Number of ejected photo electrons increase with increase
