Lights of two different frequencies whose photons have energies 1 and 2.5 eV, respectively, successively illuminate a metal whose work function is 0.5 eV. The ratio of the maximum speeds of the emitted electrons

Options

A. `1:5`
B. `1:4`
C. `1:2`
D. `1:1`

Answer

C

Explanation

Photoelectric equation gives
`KE=hv-phi_(0)`
`implies(KE)_(1)=1-0.5=0.5eV`
Similarly, `(KE)_(2)=2.5-0.5=2eV` ltBrgt `therefore(((KE))_(1))/(((KE))_(2))=(1)/(4)`
Since, `((1)/(2)mv_(1)^(2))/((1)/(2)mv_(2)^(2))=(1)/(4)`
`implies(v_(1)^(2))/(v_(2)^(2))=(1)/(4)implies(v_(1))/(v_(2))=(1)/(2)`