1.

The K.E. of the electron is E, when the incident wavelength is lamda. To increase the K.E. of the electron to 2E, the incident wavelength must be :

Answer»

`(HC)/(Elamda-hc)`
`(hclamda)/(Elamda+hc)`
`(hlamda)/(Elamda+hc)`
`(hclamda)/(Elamda-hc)`

Solution :`(hc)/(LAMBDA)=hv_(0)+E and (hc)/(X)=hv_(0)+2E`
`or (hc)/(lambda)-E=(hc)/(x)-2E,` SOLVING `x=(hc lambda)/(lambda+hc)`


Discussion

No Comment Found

Related InterviewSolutions