Q. difference between and occupied level of Mercury and and unoccupied higher level is 5 electron volt and electron having energy of five electron volt passes through Mercury vapour find the wavelength of emitted radiation?.
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lemda = hc/ E
= 6.625× 10^-34 ×3 × 10^8/ 5 × 1.6 × 10^-19
= 250 nm
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Answered by
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Answer:
wavelenght(lambda)= hc/e
=>wavelength of mercury=6.625×10-34 ×3/5 ev
=>wave length=6.625×10^-34/5×1.6×10^-19
=> lambda= 250 nm.
hope this helps
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