Question

The maximum velocity of photoelectron emitted is 4.8 m/s. If the e/m ratio of the electron is 1.76 × 10¹¹ C/kg, then stopping potential is given by ______ 

पर्याय

• 3.5 x 10^-10 J/C

• 5.5 x 10^-7 J/C

• 6.5 x 10^-11 J/C 

• 7.5 x 10² J/C

Answer 1

The maximum velocity of photoelectron emitted is 4.8 m/s. If the e/m ratio of the electron is 1.7 6 × 10¹¹ C/kg, then stopping potential is given by 6.5 x 10^-11 J/C.

Explanation:

eV₀ = `1/2"mv"_"max"^2"`

∴ V₀ = `"mv"_"max"^2/(2"e") = ("v"_"max"^2)/(2("e""/""m")) = (4.8)^2/(2 xx 1.76 xx 10^11)`

= 6.5 × 10^-11 J/C