Answer:
Wavelength of the first four spectral line:




Explanation:
Lman series when electron from higher shell number and jumps to 1st shell then all the possible lines is called as lyman series.
to calculate wavelength of first four spectral line:
For hydrogen Z=1;
by using rydberg equation
![\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3DRZ%5E2%5B%5Cfrac%7B1%7D%7Bn_1%5E2%7D%20-%5Cfrac%7B1%7D%7Bn_2%5E2%7D%20%5D)
1. n=2 to n=1
![\frac{1}{\lambda} =RZ^2[\frac{1}{n_1^2} -\frac{1}{n_2^2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3DRZ%5E2%5B%5Cfrac%7B1%7D%7Bn_1%5E2%7D%20-%5Cfrac%7B1%7D%7Bn_2%5E2%7D%20%5D)
=rydberg constant


2. n=3 to n=1


3. n=4 to n=1


4. n=5 to n=1


The correct answer would be air
The name is Potassium bromide.
Answer:
D
Explanation:
because in a decay series a daughter nuclei may be stable or decay itself. that starts a decay series