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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)

Grade:12

2 Answers

Amarjit
105 Points
3 years ago
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
15 Points
14 days ago
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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