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Solve the question 28. "e (Z make from the fifth to the fourth orbit and are incident normally on a metal plate and Photo by the shorter wavelength is or the potential the photoelectrons ejected • 1.094 O ' mA ) A •on a — 6 two successive trmsitions and ruches the ground state. In • I cv is emitted Fed he of the photon emitted in the sec«xi transition. the Of n in the intertrrdiate state? A gs of &ntical h"hogen life atoms has same atoms the lowest alerß level A and some kvel B and there are no atoms in any other energy level. The atoms of the to a hig'rr energy level by absorbing monochromatic light of photons energy 1.56 ey Sub-.equent'y, the emit radiation of only six different photon energies. Some of emitted photons enerW 7. SO eVs some have more and sorne have less than 7.56 eV_ Calculate (i) the pnnclpal number of the initially excited level B, 'mum arul minimum energies of emitted phomsand toatsat'on for gas atoms. ne longest wavelength in the Lyman series is 12 1 _ 5 nm and the shortest wavelength in the Balmerseries is 364.6 am. Use the figures to find the longest wavelength of light that could ionize hydrogen. An excited hydrogen atom emits a photon of wavelength in returning to the ground state. Denve a formula that gives the quantum number ofthe initial excited state in terms of). and R. Use this fonnulato find ni fora 102.55-nm photon A beam ofmonochromatic light of wavelength ejecb*rotoelectrons from a cesium surface I .9eV)• These photoelectrons are made to collide with hydrogen atoms in ground state. Find the Of which (a) hydrogen atoms may be ionized- (b) hydrogen atom get excited from the ground state to the first excited state and (c) the excited hy&ogen atoms may emit visible ligåL ns emitted from a metal surface of work function 1.7 eV 31. 38 9.

Dear Student, 

In lyman series, the longest wavelength is obtained when hydrogen get excited from n = 1 to n = 2. and in case of Balmer series, shortest wavelength corresponds to a transition from n = 2 to n = infinity. 
If we combine both the points, we will have the following relation 
For Lyman series and Balmer series, we have  1λL=Rh(112-122) and 1λB=Rh(122-12)
Hence Transition will be 1λfinal=Rh(112-12) = 1λL+1λB= 91.13 nm 
where λL=121.5 nm and λB = 364.6 nm 

Please let me know, if you have any other query 


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