If we express the energy of a photon in $KeV$ and the wavelength in angstroms, then energy of a photon can be calculated from the relation
- A$E = 12.4\,h\nu $
- B$E = 12.4\,h/\lambda $
- C$E = 12.4\,/\lambda $
- D$E = h\nu $
Similar Questions
The incident photon involved in the photoelectric effect experiment.
The incident photon involved in the photoelectric effect experiment.
The momentum of a photon with energy $20\, eV$ is
The momentum of a photon with energy $20\, eV$ is
The eye can detect $5 ×10^4$ photons per square metre per sec of green light ($\lambda$ $= 5000\ \mathop A\limits^o $) while the ear can detect ${10^{ - 13}}\,(W/{m^2})$. The factor by which the eye is more sensitive as a power detector than the ear is close to
The eye can detect $5 ×10^4$ photons per square metre per sec of green light ($\lambda$ $= 5000\ \mathop A\limits^o $) while the ear can detect ${10^{ - 13}}\,(W/{m^2})$. The factor by which the eye is more sensitive as a power detector than the ear is close to
$(a)$ Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500\;V$ with respect to the emitter. Ignore the small inttial speeds of the electrons. The specific charge of the electron, $i.e.$, the $e / m$ is glven to be $1.76 \times 10^{11}\; C\; kg ^{-1}$
$(b)$ Use the same formula you employ in $(a)$ to obtain electron speed for an collector potential of $10 \;MV$. Do you see what is wrong? In what way is the formula to be modified?
$(a)$ Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500\;V$ with respect to the emitter. Ignore the small inttial speeds of the electrons. The specific charge of the electron, $i.e.$, the $e / m$ is glven to be $1.76 \times 10^{11}\; C\; kg ^{-1}$
$(b)$ Use the same formula you employ in $(a)$ to obtain electron speed for an collector potential of $10 \;MV$. Do you see what is wrong? In what way is the formula to be modified?
A photon falls through a height of $1 \,km$ through the earth's gravitational field. To calculate the change in its frequency, take its mass to be $h v / c^{2}$. The fractional change in frequency $v$ is close to
A photon falls through a height of $1 \,km$ through the earth's gravitational field. To calculate the change in its frequency, take its mass to be $h v / c^{2}$. The fractional change in frequency $v$ is close to
- [KVPY 2019]