An important spectral emission line has a wavelength of $21 cm$. The corresponding photon energy is

$(h = 6.62 \times {10^{ - 34}}Js;\;\;c = 3 \times {10^8}m/s)$

  • A

    $5.9 \times {10^{ - 4}}eV$

  • B

    $5.9 \times {10^{ - 6}}eV$

  • C

    $5.9 \times {10^{ - 8}}eV$

  • D

    $11.8 \times {10^{ - 6}}eV$

Similar Questions

If the energy of a photon corresponding to a wavelength of $6000 \mathring A$ is $3.32 \times {10^{ - 19}}J$, the photon energy for a wavelength of $4000 \mathring A$ will be ............ $eV$ 

If a source of power $4\,kW$ produces $10^{20}$ photons/second, the radiation belongs to a part of the spectrum called

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

The work function of a substance is $3.0\ \mathrm{eV}$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

  • [JEE MAIN 2024]

The beam of light has three wavelengths $4144 \,\mathring A$, $4972 \;\mathring A$ and $6216\; \mathring A$ with a total intensity of $3.6 \times$ $10^{-5}\,Wm ^2$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1\,cm ^2$ of a clean metallic surface of work function $2.3\,eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2\,s$.