Question

Two photons of energies twice and thrice the work function of a metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively is –

• A. $\sqrt { 2 } : 1$
• B. $\sqrt { 3 } : 3$
• C. $\sqrt { 3 } : \sqrt { 2 }$
• D. $1 : \sqrt { 2 }$

Answer 1

Answer: (D)

$1 : \sqrt { 2 }$

Answer 2

KE = $\frac { 1 } { 2 }$ mv² = E – φ

$\frac { 1 } { 2 } \mathrm { mv } _ { 1 } ^ { 2 }$ = 2φ – φ

= φ …(1)

= 3 φ – φ = 2φ

= 2φ …(2)