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Find out the energy of H atom in the first excitation state .The value of permittivity factor `4pi epsilon_(n) = 1.11264 xx 10^(-10) C^(2)N^(-1)m^(-1)` |
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Answer» In MKS system `E_(n) = (2pi^(2)Z^(2)me^(2))/((4 piepsilon_(0))^(2)n^(2)h^(2))` `= (2 xx (3.14)^(2) xx(1)^(2) xx 9.108 xx 10^(-31) xx (1.602 xx 10^(-19))^(4))/((1.11264 xx 10^(-10))^(2) xx (2)^(2) xx (6.625 xx 10^(-34))^(2))` `= 5.443 xx 10^(-19) J` |
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