Question

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

Answer 1

⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐

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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

Answer 2

Radhe Radhe ☺️

here is your answer.......

L͟A͟Y͟M͟E͟N͟ S͟E͟R͟I͟E͟S͟:-

This is a series in which all the lines correspond to orbit having of electrons from a higher excited state to orbit having n=1 i;e n1=1, n2=2,3,4.......

see the fig.no(1)

where R is the Rydberg's constant

now see the fig. no(2)...

in fig no.2 we get lamda 1 = 4/3R

but in the above question lamda1 is given I;e

=1216A°

BAlmer series :-
----------------------

This is a series in which all the lines correspond to transition of electrons from higher excited to the orbit having n=2

see the fig.no(3)

from the above two equation........

we get....

see fig.no(4)

we know the value of lambda1 ,so we put the value of lambda1 i;e 1216A°

finally ur answer is 4864A°

HOPE IT WILL HELP YOU

$<marquee>$ By SILU12❤️

be BRAINLY☺️