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The wavelengths of photons emitted by el...

The wavelengths of photons emitted by electron transition between two similar leveis in H and `He^(+)` are `lambda_(1)` and `lambda_(2)` respectively. Then :-

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Hydrogen (H) , deuterium (D) , singly ionized helium (He^(+)) and doubly ionized lithium (Li) all have one electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelength of emitted radiations are lambda_(1), lambda_(2), lambda_(3) and lambda_(4) respectively. then approximately

A dye absorbs a photon of wavelength lambda and re - emits the same energy into two phorons of wavelengths lambda_(1) and lambda_(2) respectively. The wavelength lambda is related with lambda_(1) and lambda_(2) as :

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Calculate the wavelength of a photon emitted when an electron in H- atom maker a transition from n=2 to n=1

Calculate the wavelength of a photon emitted when an electron in H- atom maker a transition from n=2 to n=1

In a set of experiment on a hydrogen on a hypotherical one-electron atom, the wavelength of the photons emitted from transition ending in the ground state (n = 1) are shown in the energy level diagram lambda_(5 to 1) = 73.86 nm lambda_(4 to 1) = 75.63 nm lambda_(3 to 1) = 79.76 nm lambda_(2 to 1) = 94.54 nm If an electron made a transition from n = 4 to n = 2 level, the wavelength of the light that it would emit is nearly

In a set of experiment on a hydrogen on a hypotherical one-electron atom, the wavelength of the photons emitted from transition ending in the ground state (n = 1) are shown in the energy level diagram lambda_(5 to 1) = 73.86 nm lambda_(4 to 1) = 75.63 nm lambda_(3 to 1) = 79.76 nm lambda_(2 to 1) = 94.54 nm If an electron made a transition from n = 4 to n = 2 level, the wavelength of the light that it would emit is nearly