If the momentum of an electron is changed by P, then de-Broglie wavelength associated with it
changes by 0.2%. The initial momentum of electron will be about :
In photoelectric effect, the curve between photoelectric current and anode potential V (for
different frequencies) is shown in figure, then :
When the energy of the incident radiation is increased by 20%, the kinetic energy of the
photoelectrons emitted from a metal surface increased from 0.5eV to 0.8eV. The work
function of the metal is :
The de Broglie wavelength of an electron moving with a velocity
\[1.5\times 10^{8}m/s\] is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is :
A 200 W sodium street lamp emits yellow light of wavelength 0.6Ξm. Assuming it to be 25%
efficent converting electrical energy to light, the number of photons of yellow light it emits per
second is :
The ratio of wavelength of deutron and proton accelerated through the same potential difference will be :
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage
change in the de-Broglie wavelength of the particle is :
The maximum kinetic energy of the emitted photoelectrons in photoelectric effects is independent of:
According to Einstein's photoelectric equation, \(K_{\text{max}} = h\nu - \phi\). The maximum kinetic energy depends on the frequency/wavelength of the incident light and the work function, but is independent of the intensity of the light.
The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:
The de Broglie wavelength for an electron accelerated through a potential \(V\) is given by \(\lambda = \frac{1.227}{\sqrt{V}}\text{ nm}\). Substituting \(V = 81\text{ V}\), we get \(\lambda = \frac{1.227}{9}\text{ nm} \approx 0.136\text{ nm}\).
The number of photons per second on an average emitted by the source of monochromatic light of wavelength \(600\text{ nm}\), when it delivers the power of \(3.3 \times 10^{-3}\text{ watt}\) will be (\(h = 6.6 \times 10^{-34}\text{ J s}\))
Power \(P = n \frac{hc}{\lambda}\), where \(n\) is the number of photons emitted per second. Substituting the given values, we get \(n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}\text{ s}^{-1}\).