1.

Wavelength of first line of lyman series is 1216a°the wavelength of second member of balmer series is ?

Answer»

⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐


___________________________


Hey mate!


First member of lyman series(2---1)


Also,


1/lembda = RZ^2(1/n1^2 -1/n2^2)


Where,


R =1/912A°


Lembda given =1216A°


So,


1/1216A° =1/912A°Z^2(1/1^2 -1/2^2)


3/4 =Z^2(3/4)


Z =1


Now,


Second member of BALMAR series(4---2)


Putting the value of

Z =1


In the above formula..

You will get..


Lembda =4864A°


Here..n1=2

And n2=4


Hope it will HELP you..


#phoenix



Discussion

No Comment Found

Related InterviewSolutions