Question

Figure represents a graph of kinetic energy of most energetic photoelectrons Km (in eV) and frequency $\upsilon$ for a metal used as cathode in photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is

• A. $1\times {10}^{14}Hz$
• B. $1.5\times {10}^{14}Hz$
• C. $2\times {10}^{14}Hz$
• D. $2.7\times {10}^{14}Hz$

Answer 1

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

From graph, $\upsilon ={10}^{15}Hz$,

$k_{max}=3eV=3\times 1.6\times {10}^{-19}J$

According to Einsteins photoelectric equation

$k_{max}=h\upsilon -h{\upsilon }_0$

or $h{\upsilon }_0=h\upsilon -k_{max}$

or ${\upsilon }_0=\upsilon -\frac{k_{max}}{h}$ $={10}^{15}-\frac{3\times 1.6\times {10}^{-19}}{6.63\times {10}^{-34}}$

${\upsilon }_0=(10-7.3)\times {10}^{14}=2.7\times {10}^{14}Hz$