1.

The potential energy of the orbital electron i the ground state of hydrogen atom is -E. What is its kinetic energy?

Answer»

`(E)/(4)`
`(E)/(2)`
2E
4E

Solution :For ground state of `H_(2)`
`(MV^(2))/(r)=(KE^(2))/(r^(2)) RARR (1)/(2) mv^(2)=(ke^(2))/(2r)...(1)`
Potential energy `=-(ke^(2))/(r)....(2)`
Say E is potential energy then `E=-(ke^(2))/(r)`
`:.` Kinetic energy `=-(ke^(2))/(r)=(1)/(2)(-(ke^(2))/(r))`
`:.`Kinetic energy `=(E)/(2)`


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