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The work function of tungsten is4. 50ev.The wavelength of fastest electron emiited when light whose photon energy is 5.50ev falls on tungsten surfaces?
a) 12.27
b) 0.286
c) 12400
d) 1.227?
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Most Upvoted Answer
The work function of tungsten is4. 50ev.The wavelength of fastest elec...
HF=KE + H•f•
HC/wavelength=5.5ev+4.5ev
6.626×10^-34Js×3×10^8m/s÷wavelength=10ev×1.6×10-19J
wavelength=(19.89÷16)10^-7m
wavelength=124nm=124000pm
Community Answer
The work function of tungsten is4. 50ev.The wavelength of fastest elec...
Given:
- Work function of tungsten (φ) = 4.50 eV
- Photon energy of incident light (E) = 5.50 eV

To find:
- The wavelength (λ) of the fastest electron emitted from the tungsten surface.

Solution:

The energy of a photon is given by the equation:

E = hc/λ

Where:
- E is the energy of the photon
- h is the Planck's constant (6.626 x 10^-34 J.s)
- c is the speed of light (3 x 10^8 m/s)
- λ is the wavelength of the photon

We can rearrange the equation to solve for λ:

λ = hc/E

Step 1: Convert the given photon energy from eV to Joules.
1 eV = 1.6 x 10^-19 J

E = 5.50 eV * (1.6 x 10^-19 J/eV)
E = 8.8 x 10^-19 J

Step 2: Substitute the values of h, c, and E into the equation for λ.

λ = (6.626 x 10^-34 J.s * 3 x 10^8 m/s) / (8.8 x 10^-19 J)

Simplifying the equation gives:

λ = 2.25 x 10^-6 m

Step 3: Convert the wavelength from meters to angstroms.
1 meter = 1 x 10^10 angstroms

λ = 2.25 x 10^-6 m * (1 x 10^10 angstroms/m)
λ = 22,500 angstroms

Therefore, the wavelength of the fastest electron emitted from the tungsten surface is 22,500 angstroms.

Answer:
The correct option is (c) 12.27.
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The work function of tungsten is4. 50ev.The wavelength of fastest electron emiited when light whose photon energy is 5.50ev falls on tungsten surfaces? a) 12.27b) 0.286c) 12400d) 1.227?
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