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Answers
QUESTION :-
A photon of wavelength 4 × 10⁻⁷ m strikes on a metal surface , the work function of the metal being 2.13 eV . ( 1eV = 1.6020× 10⁻¹⁹ J ).
Calculate-
( i ) The energy of the photon( eV )
( ii ) The kinetic energy of the emission
( iii ) The velocity of the photo electron
SOLUTION :-
GIVEN -
- Wavelength ( λ ) = 4 × 10⁻⁷ m = 4× 10⁻⁷ × 10⁹ nm = 400nm
→ [ 1 m = 10⁹ nm ]
- Work function ( Φ ) = 2.13 eV
TO FIND -
- The energy of the photon ( eV )
Energy = hc / λ
where,
→ h = planck's constant = 6.636 × 10⁻³⁴ J-sec = 4.135× 10⁻¹⁵eV-sec
→ c = speed of light in vacuum = 3 × 10⁸ m/sec
Energy = ( 6.636 * 10⁻³⁴ J- sec )×( 3 * 10⁸ m/sec ) / 4 * 10⁻⁷ m
⇒ Energy = ( 19.908 * 10⁻²⁶ / 4 * 10⁻⁷ ) J
⇒ Energy = 4.977 * 10⁻¹⁹ J
⇒ Energy = (4.977 * 10⁻¹⁹ J) / ( 1.6020 * 10⁻¹⁹ J eV⁻¹ )
⇒ Energy = 3.106 eV
∴ The energy of the photon is 3.106 eV
2. The kinetic energy of the emision
Energy = Work Function + Kintetic Energy
→ Kinetic energy = Energy - Work function
→ Kinetic energy = 3.106 eV - 2.13 eV
→ Kinetic energy = 0.976 eV
→ Kinetic energy = 0.976 × 1.6020* 10⁻¹⁹ J
→ Kinetic energy = 1.56 * 10⁻¹⁹ J
∴ The kinetic energy of the emission is 1.56 * 10⁻¹⁹ J
3. The velocity of the photo electron
Kinetic Energy = 1/2 * m * v²
where,
m = mass of the photo electron = 9.1* 10⁻³¹ kg
v = velocity of the photo electron
→ 1.56 * 10⁻¹⁹ J = 1/2 * ( 9.1× 10⁻³¹ kg ) * v²
→ v² = (1.56 × 2 × 10 ⁻¹⁹ J ) / ( 9.1 × 10⁻³¹ kg )
→ v² = 0.343 × 10¹² m² sec⁻²
→ v = √ ( 0.343 * 10¹² m² sec⁻² )
→ v = 0.5856 * 10⁶ m sec⁻¹
∴ The velocity of the photo electron is 0.5856 * 10⁶ m sec⁻¹