# The ionization enegry of the electron in the hydrogen atom in its ground state is 13.6 ev. The atoms are excited to higher energy levels to emit rad

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The ionization enegry of the electron in the hydrogen atom in its ground state is 13.6 ev. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
A. n = 4 to n = 3 states
B. n = 3 to n = 2 states
C. n = 3 to n = 1 states
D. n = 2 to n = 1 states

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(n(n-1))/(2)=6
implies n=4
For maximum wavelength energy difference betweeen states should be minimum because
lambda=(hc)/(DeltaE)
So , transition state in n=4 to n=3