Question

A photon of wavelength 4×10 −7 m strikes on metal surface, the work function of the metal being 2.13eV. Calculate (i) the energy of the phpton (eV), (ii) the kinetic energy of the emission, (iii) the velocity of the photoelectron. (1eV=1.6020×10 −19 J).

Answer 1

$\checkmark$ Given:

A photon of wavelength 4×10⁻⁷ m strikes on metal surface, the

work function of the metal being 2.13 eV

( 1 eV=1.6020×10⁻¹⁹ J )

$\checkmark$ To Find:

1. Energy of the photon (eV)
2. Kinetic energy of the emission
3. Velocity of the photo-electron

$\checkmark$ Solution:

$\orange{1.}$

We know that,

$\red{\bigstar \ \boxed{\sf E=h \nu=\dfrac{hc}{\lambda}}}$

Here,

• E = Energy of photon
• h = planck's constant
• c = speed of light
• λ = wavelength

According to the Question,

We are asked to find the energy of photon

Given that,

A photon of wavelength 4×10⁻⁷ m strikes on metal surface, the work

function of the metal being 2.13 eV

Hence,

• λ = 4 x 10⁻⁷ m

We know that,

• h = 6.63 x 10⁻³⁴ J s
• c = 3 x 10⁸ m/s

Substituting the values,

We get,

$\blue{\sf \implies E=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4 \times 10^{-7}} \ J}$

On further simplification,

We get,

$\green{\sf \implies E=4.97 \times 10^{-19} \ J}$

According to the Question,

We are asked to give the answer in eV

Therefore,

$\orange{\sf \implies E=\dfrac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \ eV}$

On further simplification,

We get,

$\pink{\sf \implies E=3.1 \ eV}$

Hence,

$\star$ Energy of the photon = 3.1 eV

$\orange{2.}$

We know that,

$\red{\bigstar \ \boxed{\sf K.E=h \nu-h \nu_{0}}}$

Here,

• K.E = Kinetic Energy
• $\sf h \nu$ = energy of photon
• $\sf h \nu_{0}$ = work-function of metal

According to the Question,

We are asked to find the kinetic energy of emission

Given that,

A photon of wavelength 4×10⁻⁷ m strikes on metal surface, the

work function of the metal being 2.13 eV

Hence,

• $\blue{\sf h \nu_{0}=2.13 \ eV}$

We found out that,

• $\green{\sf h \nu=3.1 \ eV}$

Substituting the values,

We get,

$\orange{\sf \implies K.E=3.1 \ eV-2.13 \ eV}$

$\pink{\sf \implies K.E=0.97 \ eV}$

Hence,

$\star$ Kinetic energy = 0.97 eV

$\orange{3.}$

We know that,

$\red{\bigstar \ \boxed{\sf K.E=\dfrac{1}{2}mv^{2}}}$

Here,

• K.E = Kinetic Energy
• m = mass of photo-electron
• v = velocity of photo-electron

According to the Question,

We are asked to find the velocity of the photo-electron

We know that,

• m = 9.11 x 10⁻³¹

We found out that,

• K.E = 0.97 eV

Since,

$\longrightarrow e=1.6 \times 10^{-19}$

Conversion:

$\longrightarrow \rm 0.97 \ eV=0.97 \times 1.6 \times 10^{-19} \ V$

Substituting the values,

We get,

$\blue{\sf \implies 0.97 \times 1.6 \times 10^{-19}=\dfrac{1}{2} \times 9.11 \times 10^{-31} \times v^{2}}$

On further simplification,

We get,

$\orange{\sf \implies v^{2}=34.09 \times 10^{10} \ m/s}$

By Approximation,

We get,

$\blue{\sf \implies v^{2} \approx 34.1 \times 10^{10} \ m/s}$

Therefore,

$\green{\sf \implies v=\sqrt{34.1 \times 10^{10}} \ m/s}$

Taking Square-root

We get,

$\pink{\sf \implies v=5.84 \ m/s}$

Hence,

$\star$ Velocity = 5.84 m/s

Finally,

$\red{\rm \checkmark \ Energy = 3.1 \ eV}$

$\orange{\rm \checkmark \ Kinetic \ Energy = 0.97 \ eV}$

$\pink{\rm \checkmark \ Velocity = 5.84 m/s}$