I found an answer from ncert.nic.in

Chapter 11.pmd

This unit of energy is commonly used in atomic and nuclear physics. The work function ... 11.5 PHOTOELECTRIC EFFECT AND WAVE THEORY. OF LIGHT ... photon. Equation (11.2) is known as Einstein's photoelectric equation. We now see ... The dual (wave-particle) nature of light (electromagnetic radiation, in general) ...

For more information, see Chapter 11.pmd

I found an answer from els2017.arc.nasa.gov

template for two-page abstracts in Word 97 (PC)

Apr 10, 2017 ... We are hoping the ELS program will find your interest and that you will ... Wilhelms-Universität Münster, Germany in partnership with NASA's ... METAL-‐ SILICATE PARTITIONING OF K AS A FUNCTION ... rays and solar wind for future human missions. ... OHB System as part of the Lunar Volatiles Mobile.

For more information, see template for two-page abstracts in Word 97 (PC)

Given that

Wavelength of Ultraviolet light \lambda = 2271 Angstrom = 2271 * 10^{-10} m

Power of Ultraviolet light = 100 W

Stopping potential V = - 1.3 volts

Wavelength of red light \lambda _r = 6328 \text{ Angstrom } = 6328 * 10^{-10} m

Intensity = 10^5 W m^{-2}

Step 1: Get an expression for the work function

             We know that

                By conservation by law, Energy E = eV...........................(1)

                     Where, e - electric charge and V - potential of electrons  

                 Einstein's photoelectric equation

                            Energy of proton  E=h\upsilon\ -\ \phi

                                                         eV = h \frac{c}{\lambda} - \phi                              \because \text{ frequency } \upsilon = \frac{c}{\lambda}   

                                                         \phi = h \frac{c}{\lambda} - eV      

Step 2: Evaluating the work function of the metal

                Planck's constant, h = 6.63* 10^{-34} J

                Electron charge, e = 1.6 * 10^{-19} C

                     \phi = 6.63* 10^{-34} * \frac{3* 10^{8}}{2271 * 10^{-10}} - (1.6 * 10^{-19}) * (1.3)

                     \phi = 8.72 *10^{-19} - 2.08 *10^{-19}

                     \phi = 6.64 * 10^{-19}   Joules

                     \phi = 4.15 eV                 \because 1 eV = 1.6*10^{-19}

                Therefore, work function of the metal   \phi = 4.15 eV

Step 3: Calculating the threshold frequency of the metal  

              Relation between the work function and threshold frequency

                       \phi = h \upsilon

              Threshold frequency \upsilon = \frac{\phi}{h} \phi = h \upsilon

                       \upsilon = \frac{6.64 * 10^{-19}}{6.63* 10^{-34}}   

                       \upsilon = 1.0006 * 10^{15} Hz

            Hence, threshold frequency \upsilon = 1.0006 * 10^{15} Hz

Step 4: Frequency of the red light estimation

                   \lambda _r = 6328 \text{ Angstrom } = 6328 * 10^{-10} m

                Frequency of the red light \upsilon_r = \frac{c}{\lambda_r}   

                         \upsilon_r = \frac{ 3* 10^8}{6328 * 10^{-10}}

                         \upsilon_r = 4.74 * 10^{14} Hz

                Thus, frequency of the red light \upsilon_r = 4.74 * 10^{14} Hz

                Therefore, The photocell will not react to this red light, however strong its intensity may be. since frequency of the red

                light is less than threshold frequency (   \upsilon > \upsilon_r )