A metal is irradiated with a light of wavelength 660 nm . Given that the work function of the metal is 1.0 eV , the de-Broglie wavelength of the ejected electron is close to

Question

A metal is irradiated with a light of wavelength 660 nm . Given that the work function of the metal is 1.0 eV , the de-Broglie wavelength of the ejected electron is close to (A) 6.6× 10- 7m (B) 8.9× 10- 11m (C) 1.3× 10- 9m (D) 6.6× 10- 33m

Tardigrade Admin · Accepted Answer

We know the kinetic energy of the electron is given by:
K.E.

= E_{p} - ϕ K . E . = \frac{6.6 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{660 \times 1 0 ^{- 9}} J - 1.6 \times 1 0^{- 19} J K .

E . = 1.4 \times 1 0^{- 19} J

And the de-broglie wavelength is given by:

λ = \frac{h}{2 m K . E .} λ = \frac{6.6 \times 1 0 ^{- 34}}{2 \times 9.1 \times 1 0 ^{- 31} \times 1.4 \times 1 0 ^{- 19}} = 1.3 \times 1 0^{- 9} m

Q. A metal is irradiated with a light of wavelength $660 nm$ . Given that the work function of the metal is $1.0 e V$ , the de-Broglie wavelength of the ejected electron is close to

$6.6 \times 1 0^{- 7} m$

$8.9 \times 1 0^{- 11} m$

$1.3 \times 1 0^{- 9} m$

$6.6 \times 1 0^{- 33} m$

Q. A metal is irradiated with a light of wavelength 660nm . Given that the work function of the metal is 1.0eV , the de-Broglie wavelength of the ejected electron is close to

6.6×10−7m

8.9×10−11m

1.3×10−9m

6.6×10−33m

We know the kinetic energy of the electron is given by: K.E. =Ep​−ϕK.E.=660×10−96.6×10−34×3×108​J−1.6×10−19JK. E.=1.4×10−19J And the de-broglie wavelength is given by: λ=2mK.E.​h​λ=2×9.1×10−31×1.4×10−19​6.6×10−34​=1.3×10−9m

Q. A metal is irradiated with a light of wavelength $660 nm$ . Given that the work function of the metal is $1.0 e V$ , the de-Broglie wavelength of the ejected electron is close to

$6.6 \times 1 0^{- 7} m$

$8.9 \times 1 0^{- 11} m$

$1.3 \times 1 0^{- 9} m$

$6.6 \times 1 0^{- 33} m$