307409 How many photons of light having a wavelength of \({\rm{4000}}\,\mathop {\rm{A}}\limits^{\rm{^\circ }} \) are necessary to provide 1 J of energy? \({\rm{(h = 6}}{\rm{.63 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{J}}{\rm{.s,c = 3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{m/s)}}\)

307410 A near \(\mathrm{UV}\) photon of \(300 \mathrm{~nm}\) is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength \(760 \mathrm{~nm}\). Hence, wavelength of the second photon is

307411 The energies \({{\rm{E}}_{\rm{1}}}\) and \({{\rm{E}}_{\rm{2}}}\) of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., \({{\rm{\lambda }}_{\rm{1}}}\) and \({{\rm{\lambda }}_{\rm{2}}}\) will be

307412 A bulb emits electromagnetic radiation of \(\mathrm{660 \mathrm{~nm}}\) wavelength. The total energy of radiation is \(\mathrm{3 \times 10^{-18} \mathrm{~J}}\). The number of emitted photons will be
\(\mathrm{\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \times \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}\)

307409 How many photons of light having a wavelength of \({\rm{4000}}\,\mathop {\rm{A}}\limits^{\rm{^\circ }} \) are necessary to provide 1 J of energy? \({\rm{(h = 6}}{\rm{.63 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{J}}{\rm{.s,c = 3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{m/s)}}\)

307410 A near \(\mathrm{UV}\) photon of \(300 \mathrm{~nm}\) is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength \(760 \mathrm{~nm}\). Hence, wavelength of the second photon is

307411 The energies \({{\rm{E}}_{\rm{1}}}\) and \({{\rm{E}}_{\rm{2}}}\) of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., \({{\rm{\lambda }}_{\rm{1}}}\) and \({{\rm{\lambda }}_{\rm{2}}}\) will be

307412 A bulb emits electromagnetic radiation of \(\mathrm{660 \mathrm{~nm}}\) wavelength. The total energy of radiation is \(\mathrm{3 \times 10^{-18} \mathrm{~J}}\). The number of emitted photons will be
\(\mathrm{\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \times \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}\)

307409 How many photons of light having a wavelength of \({\rm{4000}}\,\mathop {\rm{A}}\limits^{\rm{^\circ }} \) are necessary to provide 1 J of energy? \({\rm{(h = 6}}{\rm{.63 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{J}}{\rm{.s,c = 3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{m/s)}}\)

307410 A near \(\mathrm{UV}\) photon of \(300 \mathrm{~nm}\) is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength \(760 \mathrm{~nm}\). Hence, wavelength of the second photon is

307411 The energies \({{\rm{E}}_{\rm{1}}}\) and \({{\rm{E}}_{\rm{2}}}\) of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., \({{\rm{\lambda }}_{\rm{1}}}\) and \({{\rm{\lambda }}_{\rm{2}}}\) will be

307412 A bulb emits electromagnetic radiation of \(\mathrm{660 \mathrm{~nm}}\) wavelength. The total energy of radiation is \(\mathrm{3 \times 10^{-18} \mathrm{~J}}\). The number of emitted photons will be
\(\mathrm{\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \times \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}\)

307409 How many photons of light having a wavelength of \({\rm{4000}}\,\mathop {\rm{A}}\limits^{\rm{^\circ }} \) are necessary to provide 1 J of energy? \({\rm{(h = 6}}{\rm{.63 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{\rm{J}}{\rm{.s,c = 3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{m/s)}}\)

307410 A near \(\mathrm{UV}\) photon of \(300 \mathrm{~nm}\) is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength \(760 \mathrm{~nm}\). Hence, wavelength of the second photon is

307411 The energies \({{\rm{E}}_{\rm{1}}}\) and \({{\rm{E}}_{\rm{2}}}\) of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, i.e., \({{\rm{\lambda }}_{\rm{1}}}\) and \({{\rm{\lambda }}_{\rm{2}}}\) will be

307412 A bulb emits electromagnetic radiation of \(\mathrm{660 \mathrm{~nm}}\) wavelength. The total energy of radiation is \(\mathrm{3 \times 10^{-18} \mathrm{~J}}\). The number of emitted photons will be
\(\mathrm{\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \times \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}\)