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#### Example – 01:

- The energy of a photon is 2.59 eV. Find its frequency and wavelength.
**Given:**Energy of photon = E = 2.59 eV = 2.59 x 1.6 x 10^{-19}J, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Frequency of photon = ν = ?, Wavelength = λ = ?**Solution:**

We have E = h ν

∴ ν = E/h = (2.59 x 1.6 x 10^{-19})/(6.63 x 10^{-34}) = 6.244 x 10^{14} Hz

Now c = ν λ

∴ λ = c/ν = (3 x 10^{8})/( 6.244 x 10^{14}) = 4.805 x 10^{-7} m

∴ λ = 4805 x 10^{-10} m = 4805 Å

**Ans:** The frequency of photon is 6.244 x 10^{14} Hz and its wavelength is 4805 Å

#### Example – 02:

- The energy of a photon is 1.0 x 10
^{-8}J. Find its frequency and wavelength. **Given:**Energy of photon = E = 1.0 x 10^{-18}J, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Frequency of photon = ν = ?, Wavelength = λ = ?**Solution:**

We have E = h ν

∴ ν = E/h = (1.0 x 10^{-18})/(6.63 x 10^{-34}) = 1.508 x 10^{15} Hz

Now c = ν λ

∴ λ = c/ν = (3 x 10^{8})/( 1.508 x 10^{15}) = 1.989 x 10^{-7} m

∴ λ = 1989 x 10^{-10} m = 1989 Å

**Ans:** The frequency of photon is 1.508 x 10^{14} Hz and its wavelength is 1989 Å

#### Example – 03:

- The energy of a photon is 300 eV. Find its wavelength.
**Given:**Energy of photon = E = 300 eV = 300 x 1.6 x 10^{-19}J, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Wavelength = λ = ?**Solution:**

We have E = h ν = hc/λ

∴ λ = hc/E = (6.63 x 10^{-34})(3 x 10^{8})/(300 x 1.6 x 10^{-19}) = 4.144 x 10^{-9} m

∴ λ = 41.44 x 10^{-10} m = 41.44 Å

**Ans:** The wavelengthof photon is 41.44 Å

#### Example – 04:

- Find the energy of a photon in eV if its wavelength is 10 m
**Given:**Wavelength of photopn = λ = 10 m, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Wavelength = λ = ?**Solution:**

We have E = h ν = hc/λ

∴ E = hc/λ = (6.63 x 10^{-34})(3 x 10^{8})/(10) = 19.89 x 10^{-27} J

∴ E = (19.89 x 10^{-27})/(1.6 x 10^{-19}) = 1.243 x 10^{-7} eV

**Ans:** The energy of the photon is 1.243 x 10^{-7} eV

#### Example – 05:

- Find the energy of a photon whose frequency is 5.0 x 10
^{14}Hz **Given:**Frequency of photon = ν = 5.0 x 10^{14}Hz, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Energy of photon = E = ?**Solution:**

We have E = h ν

∴ E = (6.63 x 10^{-34}) x (5.0 x 10^{14})= 3.315 x 10^{-29 } J

**Ans:** The energy of the photon is 3.315 x 10^{-29 } J

#### Example – 06:

- The photoelectric work function of silver is 3.315 eV. calculate the threshold frequency and threshold wavelength of silver.
**Given:**Work function of silver = Φ = 3.315 eV = 3.315 x 1.6 x 10^{-19}J, speed of light = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Threshold frequency of silver = ν_{o}= ?, Threshold wavelength of silver = λ_{o}= ?**Solution:**

We have Φ = h ν_{o}

∴ ν_{o} = Φ/h = (3.315 x 1.6 x 10^{-19})/(6.63 x 10^{-34}) = 8 x 10^{14} Hz

Now c = ν_{o} λ_{o}

∴ λ_{o} = c/ν_{o} = (3 x 10^{8})/( 8 x 10^{14}) = 3.750 x 10^{-7} m

∴ λ_{o}= 3750 x 10^{-10} m = 4805 Å

**Ans:** The threshold frequency of silver is 8 x 10^{14} Hz and its threshold wavelength is 3750 Å

#### Example – 07:

- A light of wavelength 4800 Å can just cause photoemission from a metal. What is the photoelectric work function for metal in eV?
**Given:**Threshold wavelength = λ_{o}= 4800 Å = 4800 x 10^{-10}m, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Work function of silver = Φ = ?**Solution:**

We have Φ = h ν_{o }= hc/λ_{o}

∴ Φ = (6.63 x 10^{-34})x(3 x 10^{8})/(4800 x 10^{-10}) = 4.144 x 10^{-19} J

∴ Φ = (4.144 x 10^{-19})/(1.6 x 10^{-19}) = 2.59 eV

**Ans:** The photoelectric work function of the metal is 2.59 eV

#### Example – 08:

- The photoelectric work function of a metal is 2 eV. calculate thelowest frequency radiation that will cause photoemission from the surface.
**Given:**Work function of silver = Φ = 2 eV = 2 x 1.6 x 10^{-19}J, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Threshold frequency of silver = ν_{o}= ?,**Solution:**

We have Φ = h ν_{o}

∴ ν_{o} = Φ/h = (2 x 1.6 x 10^{-19})/(6.63 x 10^{-34}) = 4.827 x 10^{14} Hz

**Ans:** The threshold frequency of metal is 4.827 x 10^{14}.

#### Example – 08:

- The photoelectric work function of platinum is 6.3 eV and longest wavelength that can eject photoelectron from platinum is 1972 Å. Calculate the Planck’s constant.
**Given:**Work function of platinum = Φ = 6.3 eV =6.3 x 1.6 x 10^{-19}J, Threshold wavelength of silver = 1972 Å = 1972 x 10^{-10}m, speed of light = c = 3 x 10^{8}m/s,**To Find:**Planck’s constant = h = ?,**Solution:**

We have Φ = h ν_{o }= hc/λ_{o}

∴ h = Φλ_{o}/c = (6.3 x 1.6 x 10^{-19}) x (1972 x 10^{-10})/(3 x 10^{8}) = 6.625 x 10^{-34} Js

**Ans:** The value of Palnck’s constant is 6.625 x 10^{-34} Js

#### Example – 09:

- The photoelectric work function of metal is 1.32 eV. calculate the longest wavelength that can cause photoelectric emission from the metal surface.
**Given:**Work function of silver = Φ = 1.32 eV = 1.32 x 1.6 x 10^{-19}J, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js**To Find:**Threshold wavelength of metal = λ_{o}= ?**Solution:**

We have Φ = h ν_{o }= hc/λ_{o}

∴ λ_{o} = hc/Φ =(6.63 x 10^{-34}) x (3 x 10^{8}) / (1.32 x 1.6 x 10^{-19}) = 9.418 x 10^{-7} m

∴ λ_{o}= 9418 x 10^{-10} m = 9418 Å

**Ans:** The threshold wavelength is 9418 Å

#### Example – 10:

- The photoelectric work function of metal is 5 eV. calculate the threshold frequency for the metal. If a light of wavelength 4000 Å is incident on this metal surface, will photoelectron will be ejected?
**Given:**Work function of silver = Φ = 5 eV = 5 x 1.6 x 10^{-19}J, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js, wavelength of incident light = λ = 4000 Å = 4000 x 10^{-10}m**To Find:**Threshold wavelength of metal = λ_{o}= ?**Solution:**

We have Φ = h ν_{o}

∴ ν_{o} = Φ/h = (5 x 1.6 x 10^{-19})/(6.63 x 10^{-34}) = 1.2 x 10^{15} Hz

Now c = ν λ

∴ ν = c/λ = (3 x 10^{8})/( 4000 x 10^{-10}) = 7.5 x 10^{14} Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

**Ans:** The threshold frequency is 1.2 x 10^{15} Hz and no photoelectron will be emitted.

#### Example – 11:

- The photoelectric work function of a metal is 2.4 eV. calculate the incident frequency, the threshold frequency for the metal. If a light of wavelength 6800 Å is incident on this metal surface, will photoelectron will be ejected?
**Given:**Work function of silver = Φ = 2.4 eV= 2.4 x 1.6 x 10^{-19}J, speed of light = c = 3 x 10^{8}m/s, Planck’s constant = h = 6.63 x 10^{-34}Js, wavelength of incident light = λ = 6800 Å = 6800 x 10^{-10}m**To Find:**Threshold wavelength of metal = λ_{o}= ?**Solution:**

We have Φ = h ν_{o}

∴ ν_{o} = Φ/h = (2.4 x 1.6 x 10^{-19})/(6.63 x 10^{-34}) = 5.79 x 10^{14} Hz

Now c = ν λ

∴ ν = c/λ = (3 x 10^{8})/( 6800 x 10^{-10}) = 4.41 x 10^{14} Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

**Ans:** The incident frequency is 4.41 x 10^{14} Hz and the threshold frequency is 5.79 x 10^{14} Hz,

and no photoelectron will be ejected.

#### Example – 12:

- The photoelectric work function of a metal is 3 eV. calculate the threshold frequency for the metal. If a light of wavelength 6000 Å is incident on this metal surface, will photoelectron will be ejected?
**Given:**Work function of silver = Φ = 3 eV = 3 x 1.6 x 10^{-19}J, speed of light = 3 x 10^{8}m/s, planck’s constant = h = 6.63 x 10^{-34}Js, wavelength of incident light = λ = 6000 Å = 6000 x 10^{-10}m**To Find:**Threshold wavelength of metal = λ_{o}= ?**Solution:**

We have Φ = h ν_{o}

∴ ν_{o} = Φ/h = (3 x 1.6 x 10^{-19})/(6.63 x 10^{-34}) = 7.24 x 10^{14} Hz

Now c = ν λ

∴ ν = c/λ = (3 x 10^{8})/( 6000 x 10^{-10}) = 5 x 10^{14} Hz

The frequency of incident light is less than the threshold frequency.

No photoelectrons will be emitted from the metal surface.

**Ans:** The threshold frequency is 7.24 x 10^{14} Hz,

and no photoelectron will be ejected.

Science > Physics > Electrons and Photons > You are Here |

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