Please help me with my homework! please show me howto solve this problem. I will
ID: 1722348 • Letter: P
Question
Please help me with my homework! please show me howto solve this problem.I will give proper respect as soon as possible. PLEEEASE helpme!!! #5.11 Find the range of wavelengths in the Balmer series of hydrogen.Does the Balmer series lie completely in the visible region of thespectrum? If not, what other region does it include? Answer is 365nm < =< 656nm, UV andvisible Please help me with my homework! please show me howto solve this problem.
I will give proper respect as soon as possible. PLEEEASE helpme!!! #5.11 Find the range of wavelengths in the Balmer series of hydrogen.Does the Balmer series lie completely in the visible region of thespectrum? If not, what other region does it include? Answer is 365nm < =< 656nm, UV andvisible Answer is 365nm < =< 656nm, UV andvisible
Explanation / Answer
For barmer series , n1=2 , n2 = 3,4,5,..... we know ( 1/ ) = R [ ( 1/ n1^2 ) -( 1/ n2^2 )] Take n1=1 and n2 = 3 where R = rydberg's constant = 10.96 * 10 ^ 6 m^-1 plug th evalues weget ( 1/ ) = ( 10.96 * 10 ^ 6 m ^-1 )[ ( 1/ 2 ^ 2 ) - ( 1/ 3^ 2) ] = 10.96 * 10 ^ 6 * [ ( 1/ 4 ) - ( 1/ 9 ) ] = 10.96 * 10 ^ 6 * [ 0.25 - 0.11) = 1.5222* 10 ^ 6 m ^-1 = 0.6569 * 10 ^ -6 m = 656.9 * 10 ^ -9 m = 656.9 nm For wavelength limit n1 = 2 , n2 = plug these values we get ( 1/ ) = ( 10.96 * 10 ^6m ^-1 ) [ ( 1/ 2^ 2 ) - ( 1/ ^ 2 ) ] = 10.96 * 10 ^ 6 * ( 1/ 4 ) = 2.74 * 10 ^ 6 = 0.3649 * 10 ^ -6 m = 364.9 * 10 ^ -9 m = 364.9 nm Limit of Barmer series is 364.9 nm to656.9 nm wavelength range of visible spectrum is 380 nm to750 nm 364.9 nm is lies in the infrared spectrum . So, another region is infraredRelated Questions
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