1.

For a certain metal, v is twice v_(0) and electrons come out with a maximum velocity of 4xx10^(6)ms^(-1). If value of v=5v_(0), then the maximum velocity of photoelectrons will be:

Answer»

`8xx10^(6)ms^(-1)`
`8xx10^(4)ms^(-1)`
`8xx10^(8)ms^(-1)`
`8xx10^(3)ms^(-1)`

Solution :`(1)/(2)m(4xx10^(6))^(2)=h(2v-v_(0))`
`(1)/(2)mv_("max")^(2)=h(5v_(0)-v_(0))`
`v_("max")=8xx10^(6) ms^(-1)`


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