Question

Figure represents a graph of kinetic energy of most energetic photoelectrons, $K_{max}$ (in eV), and frequency (v) for a metal used as cathode in photo electric experiment. The threshold frequency of light for the photoelectrons emission from the metal is

• A. $1\times {10}^{14}Hz$
• B. $9.5\times {10}^{14}Hz$
• C. $2.7\times {10}^{14}Hz$
• D. $2.7\times {10}^{16}Hz$

Answer 1

The correct option is C $2.7\times {10}^{14}Hz$

$hv=h{v}_0+K_{max}\Rightarrow v_0=v-\frac{K_{max}}{h}={10}^{15}-\frac{3\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}={10}^{15}-0.73\times {10}^{15}{v}_0=2.7\times {10}^{14}Hz$