Question

In an excited state of hydrogen like atom an electron has total energy of `-3.4 eV`. If the kinetic energy of the electron is E and its de-Broglie wavelength is `lambda`, then
A. `E = 6.8 eV, lambda = 6.6 xx 10^(10)`
B. `E = 3.4 eV, lambda = 6.6 xx 10^(-10) m`
C. `E = 3.4 eV, lambda = 6.6 xx 10^(-11) m`
D. `E = 6.8 eV, lambda = 6.6 xx 10^(-11) m`

Answer 1

Correct Answer - B
`P.E. = -2(K.E.)`
`T.E. = (P.E.) + (K.E.)`
`T.E. = - 2(K.E.) + (K.E.)`
`T.E. = - (K.E.) - T.E. = K.E. , K.E. = 3.4 eV`
`lambda = 6.6 xx 10^(-10) m`