Modern Physics> Ultraviolet light of wavelength Lambda 1 ...

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)

Aman Goyal , 8 Years ago

Grade 12

2 Answers

Amarjit

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.

Last Activity: 8 Years ago

Ankit Khandelwal

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

Last Activity: 4 Years ago

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Modern Physics> Ultraviolet light of wavelength Lambda 1 ...

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)

Aman Goyal , 8 Years ago

Amarjit

Ankit Khandelwal

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