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Q.

The shortest wavelength in Balmer’s series for Hydrogen atom is ...A... and this is obtained by substituting ...B ... in Balmer’s formula. Here, A and B refer to

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a

656.3 nm, n = 3

b

410.2 nm, n = 5

c

364.6 nm, n = ∞

d

486.1 nm, n = 4

answer is D.

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Detailed Solution

(d) The shortest wavelength occurs when an electron makes a transitions from n = ∞ to n = 2 state.

∴  1λmin=R122-1∞=R4

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