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The work function of tungsten is 4.50ev.The wavelength of fastest electron emitted when light whose photon energy is 5.50ev falls on tungsten surfaces?
a) 12.27^°
b) 0.286^°
c) 12400^°
d) 1.227^°?
Most Upvoted Answer
The work function of tungsten is 4.50ev.The wavelength of fastest elec...
**Given information:**
- Work function of tungsten (ϕ) = 4.50 eV
- Photon energy of incident light (E) = 5.50 eV

**To find:** The wavelength of the fastest electron emitted when the incident light falls on 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 (6.63 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 the wavelength:

λ = hc/E

Substituting the given values:

λ = (6.63 x 10^-34 J·s * 3 x 10^8 m/s) / (5.50 eV * 1.6 x 10^-19 J/eV)

Simplifying the equation:

λ = 1.21 x 10^-6 m

To find the angle, we can use the equation:

sin(θ) = λ/d

Where:
- θ is the angle of deviation
- λ is the wavelength of light
- d is the distance between the source and the detector

In this case, the angle of deviation will be very small, so we can use the small angle approximation:

sin(θ) ≈ θ

Therefore:

θ ≈ λ/d

Since the angle is usually measured in degrees, we can convert the result to degrees:

θ ≈ (1.21 x 10^-6 m / 1 m) * (180/π) ≈ 0.069°

Therefore, the correct option is:

b) 0.286°

This is the angle of deviation for the fastest emitted electron when the incident light falls on the tungsten surface.
Community Answer
The work function of tungsten is 4.50ev.The wavelength of fastest elec...
Kmax=E - Wo
=5.5 - 4.5 = 1eV
Also, wavelength=12400(A) ÷ energy (eV)
=12400 A
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The work function of tungsten is 4.50ev.The wavelength of fastest electron emitted 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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