Question

# A difference of $$2.3 eV$$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Solution

## Given: The difference in energy between the two energy levels is $$2.3eV$$.To find: The frequency of radiation emitted when the atom makes a transition between the energy levels.The energy between the energy levels is $$E=2.3eV\\\Rightarrow2.3\times1.6\times10^{-19} J$$The frequency of the radiations emitted during transition is:$$E=h\nu$$$$\nu=\dfrac hE$$$$\Rightarrow\dfrac{6.626\times10^{-34}}{2.3\times1.6\times10^{-19}}$$$$\Rightarrow5.5\times10^{14}Hz$$PhysicsNCERTStandard XII

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