Question

What transition in hydrogen spectrum would have the same wavelength in balmer transition n is equal to 42 n is equal to 2 in of h2 + spectrum?

Answer 1

Answer:

For He+,1λHe+,1λ=RHz2[122−142]=RHz2[122−142]

For H,1λH,1λ=RH[1n21−1n22]=RH[1n12−1n22]

Since λλ is same

∴Z2[122−142]=[1n21−1n22]∴Z2[122−142]=[1n12−1n22]

z=2z=2

[112−122]=[1n21−1n22][112−122]=[1n12−1n22]