Question

calculate the wavelength of light emitted when the electrons in a hydrogen atom with n =5 in balmer series​

Answer 1

Answer:

Explanation:

Given : λ=122 nm  for     n1​=1 and  n2​=2

∴        λ1​=R(1/n12​−1/n2 2​​)

⟹     1221​=R(1/1 2−1/2 2​)=43R​                             ...............(1)

Wavelength is minimum when    n1​=1 and  n2​=∞

⟹    1/ λmin​​=R(1/1 2​−1/∞)=R                                ................(2)

Dividing (1) and (2) we get: λmin/122​​=43​

⟹     λmin​=0.75×122=91.5 nm =915 Ao

Answer 2

Answer:

Explanation:

4.65e3

convert to nm