A beam of electromagnetic radiation of intensity \(6.4 \times 10^{-5} \mathrm{~W} / \mathrm{cm}^{2}\) is comprised of wavelength, \(\lambda=310 \mathrm{~nm}\). It falls normally on a metal (work function \(\varphi=2 \mathrm{eV}\) ) of surface area \(1 \mathrm{~cm}^{2}\). If one in \(10^{3}\) photons ejects an electron, total number of electrons ejected in is \(10^{\mathrm{x}} .\left(\mathrm{h_{c}}=1240 \mathrm{eVnm}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)\), then x is ____.

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Modern Physics

A beam of electromagnetic radiation of intensity

6.4 \times 1 0^{- 5} W / cm^{2}

is comprised of wavelength,

λ = 310 nm

. It falls normally on a metal (work function

φ = 2 eV

) of surface area

1 cm^{2}

. If one in

1 0^{3}

photons ejects an electron, total number of electrons ejected in is

1 0^{x} . (h_{c} = 1240 eVnm, 1 eV = 1.6 \times 1 0^{- 19} J)

, then x is ____.

Answer: 11

Each photon has energy

E = \frac{h c}{λ} = \frac{1240}{310} = 4 eV = 6.4 \times 1 0^{- 19} J

, which exceeds the work function so emission occurs. The power on the

1 cm^{2}

surface is

6.4 \times 1 0^{- 5} W

, giving a photon arrival rate of

\frac{6.4 \times 1 0 ^{- 5}}{6.4 \times 1 0 ^{- 19}} = 1 0^{14}

per second. Since only one photon in

1 0^{3}

ejects an electron, the emission rate is

\frac{1 0 ^{14}}{1 0 ^{3}} = 1 0^{11}

per second, so

x = 11

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Integer answer