UV light radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy of faster electrons emitted is approximately

3.2 \times 10^{- 15} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.2 \times 10^{- 17} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.2 \times 10^{- 19} J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.2 \times 10^{- 21} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3.2 \times 10^{- 19} J$
KE of emitted electron is
$K_{m a x} = h v - W$
= 6.2 eV - 4.2 eV
= 2 eV
$= 2 \times 1.6 \times 10^{- 19}$
$= 3.2 \times 10^{- 19} J$

Suggest Corrections

Similar questions

Q. Ultraviolet radiations of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy in joules of the fastest electron emitted is approximately

The work function of a metal is 4.2 eV. If the radiation with wavelength 2000 $0 A$ falls on the metal surface then the kinetic energy of fastest photo electron will be:

Q. Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). The kinetic energy in joule of the faster electron emitted is approximately

Q. UV light radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy of faster electrons emitted is approximately

The work function of a metal is 4.2 $e ν$ . If radiations of $2000 o A$ fallon the metal, then the kinetic energy of the fastest photoelectron is

Join BYJU'S Learning Program

UV light radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy of faster electrons emitted is approximately

The correct option is C 3.2×10−19JKE of emitted electron is Kmax=hv−W = 6.2 eV - 4.2 eV = 2 eV =2×1.6×10−19 =3.2×10−19J

The correct option is C $3.2 \times 10^{- 19} J$
KE of emitted electron is
$K_{m a x} = h v - W$
= 6.2 eV - 4.2 eV
= 2 eV
$= 2 \times 1.6 \times 10^{- 19}$
$= 3.2 \times 10^{- 19} J$