1.

A metal is irradiated with a radiation of wavelength 300 nm. Then photo electrons found to have velocity of 3.7 xx 10^5 m/s. When a graph is drawn between K.E of photo electrons and frequency of radiation a straight line is obtained cutting X-axis. Difference between incident frequency and threshold frequency is

Answer»

`9.4xx10^(13)` Hz
`8.5xx10^(22)` Hz
`9.4xx10^(12)` Hz
`8.4xx10^(10) Hz`

Solution :`upsilon - upsilon_0 =(6.625xx10^(-19))/(6.625xx10^(13)) -9.09 XX 10^(14)`
`=9.4xx10^(23) Hz`


Discussion

No Comment Found