When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectron and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively : (Assume frequency of radiation to be constant)
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 :
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 \(\lambda = \frac{1.227}{9}\text{ nm} \approx 0.136\text{ nm}\).
Ratio of shortest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is
The shortest wavelength in a series corresponds to transition from \(n_2 = \infty\) to \(n_1\). For Lyman series, \(\lambda_L = \frac{1}{R}\). For Balmer series, \(\lambda_B = \frac{4}{R}\). Therefore, the ratio \(\frac{\lambda_L}{\lambda_B} = \frac{1}{4}\).
Photocell is illuminated by a point source of light, which is placed at a distance \(d\) from the cell. If the distance becomes \(2d\), then number of electrons emitted per second will be
Intensity of light from a point source is inversely proportional to the square of the distance, \(I \propto \frac{1}{d^2}\). Since the number of photoelectrons emitted per second is proportional to intensity, doubling the distance reduces the emission to one-fourth.
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 \(n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}\text{ s}^{-1}\).