Ultraviolet light of wavelength Lambda 1 and Lambda 2 when allowed to fall on a hydrogen atoms in their Ground state is found to liberate electrons with kinetic energy 1.8ev and 4ev respectively.find the value of lambda 1/lambda 2 (L1/L2)

Question

Amarjit · Answer

Since electrons are liberated hence the total energoes must be 15.4eV and 17.6eV respectively.Now using the formula E=hc/lambda we will get lambda in both the cases keeping hc= 12400( lamba in the order of angstrom order).Now just take their ratio.

Ankit Khandelwal · Answer

Since electron travel from ground state to infinity so the energy liberated is 13.6 ev + kinetic energy 13.6 + 1.8 =15.4ev 13.6 + 4 = 17.6 ev Now we know E=hc/lamda(in angstrom) we want lambda1/lambda2= 17.6/15.4 = 8/7

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Ultraviolet light of wavelength Lambda 1 and Lambda 2 when allowed to fall on a hydrogen atoms in their Ground state is found to liberate electrons with kinetic energy 1.8ev and 4ev respectively.find the value of lambda 1/lambda 2 (L1/L2)

Ultraviolet light of wavelength Lambda 1 and Lambda 2 when allowed to fall on a hydrogen atoms in their Ground state is found to liberate electrons with kinetic energy 1.8ev and 4ev respectively.find the value of lambda 1/lambda 2 (L1/L2)

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Ultraviolet light of wavelength Lambda 1 and Lambda 2 when allowed to fall on a hydrogen atoms in their Ground state is found to liberate electrons with kinetic energy 1.8ev and 4ev respectively.find the value of lambda 1/lambda 2 (L1/L2)

Ultraviolet light of wavelength Lambda 1 and Lambda 2 when allowed to fall on a hydrogen atoms in their Ground state is found to liberate electrons with kinetic energy 1.8ev and 4ev respectively.find the value of lambda 1/lambda 2 (L1/L2)

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