The wavelength in Angstroms of the photon that is emitted when an electron in Bohr orbit n = 2 returns to the orbit n = 1 in the hydrogen atom will be: The ionization potential of the ground state of hydrogen atom is 2.17 × 10−11 erg.
A
82 nm
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B
672 nm
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C
122 nm
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D
none of these
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Solution
The correct option is C 122 nm E2 - E1 = -2.17 × 10−11 [14 - 11] erg [ n1 = 1, n2 = 2] =2.17 × 10−11×34 erg E2 - E1 = hv = cλh λ = hcΔE = 6.62×10−27erg×3.0×1010cm2.17×10−11×3/4 = 2.6482.17 10−5 cm =1.22× 10−5 cm = 1220A0