Question

A and B are two metal with threshold frequencies $1.8 \times 10 ^ { 14 } \mathrm {~Hz}$and identical photons of energy 0.825 eV each are incident on them Then photoelectrons are emitted in (Taken =6.6×10^-34Js)

• A. B alone
• B. A alone
• C. Neither A nor B
• D. Both A and B

Answer 1

Answer: (B)

A alone

Answer 2

: $\phi _ { 0 A } = \frac { h \mathrm { v } _ { 0 } } { e } \mathrm { eV } = \frac { \left( 6.6 \times 10 ^ { - 34 } \right) \times \left( 1.8 \times 10 ^ { 14 } \right) } { 1.6 \times 10 ^ { - 19 } } \mathrm { eV }$ $= 0.74 \mathrm { eV }$

$\phi _ { 0 B } = \frac { \left( 6.6 \times 10 ^ { - 34 } \right) \times \left( 2.2 \times 10 ^ { 14 } \right) } { 1.6 \times 10 ^ { - 19 } } \mathrm { eV }$ $= 0.91 \mathrm { eV }$

Since the incident energy 0.825 eVis greater than $0.74$ eV and less than so photoelectrons are emitted from metal A only