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In Lyman series, shortest wavelength of H-atom appears at x meter , then longest wavelength in Balmer series of He+ appear at
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a
9x5m
b
36x5m
c
x4m
d
5x9
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Correct option is A
For the spectral line in H-atom and H-like species (one electron) 1λ=v¯=R¯HZ21n12−1n22For Lyman seriesFor shortest wavelength (maximum wave number) n2→∞∴ 1λmin=vmax=R¯H(1)211For longest wavelength (minimum wave number), n2=n1+1For Balmer series, n1=2∴ 1λmax=vmin=RH(Z)212For He+ , Z = 21λmax=1x(2)2536⇒1λmax=59x ∴ λmax=9x5Similar Questions
At 200°C,hydrogen molecule have velocity 2.4 × 105 cm s-1. The de Broglie wavelength in this case is approximately
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