# Help with electron and radiation regarding Vmax? Does emission occur??

## What is the Vmax? the greatest velocity from an electron to emit and does emission occur at wavelength 80nm? If so, what is Vmax of the emitted electrons? threshold energy for a certain surface is 19.8x10^-19J. a) If radiation of frequency 6.3*10^15Hz is directed on to this surface, what will be the greatest velocity with which any electron could be emitted? b) If radiation of the same frequency, but twice the intensity is used, what would be Vmax? c) If radiation of wavelength of 80nm is used, does emission occur? If so, what is Vmax of the emitted electrons?

Nov 22, 2017

(a) v_(max) =2.20 × 10^6 m s⁻¹
(b) No change
(c) v_(max) = 1.05 × 10^6 m s⁻¹

#### Explanation:

(a) Equation for kinetic energy of emitted electrons:
${E}_{K \max} = h f - \phi$
And ${E}_{K \max} = \frac{1}{2} m {v}_{\max}^{2}$
$\frac{1}{2} m {v}_{\max}^{2} = h f - \phi$
⇒ v_(max) = sqrt((2(hf - phi))/m)
Substitute the values:
⇒ v_(max) = sqrt((2(6.63×10^(-34) × 6.3×10^(15) - 19.8×10^(-19)))/ 9.11×10^(-31))
⇒ v_(max) =2.20 × 10^6 m s⁻¹

(b) Intensity does not affect photon energy (hf) and therefore ${E}_{K \max}$ and ${v}_{\max}$ are unchanged.

(c) Photon energy is given by this formula:
E = hf = (hc )/λ
Substitute the wavelength variant into the equation we used for part (a).
v_(max) = sqrt((2(hc//λ - phi))/m)

Before using that equation we can do a quick check if we will get an answer for ${v}_{\max}$. If photon energy, (hc)/λ, is greater than the work function (φ) then there is emission, otherwise there is no emission.

(hc)/λ = (6.63×10^(-34) × 3.0×10^8)/ 80×10^(-9) = 2.486× 10^(-18) J
There is emission as 2.49 × 10^(-18) > 19.8 × 10^(-19)

Now substitute the values into the equation for ${v}_{\max}$:
v_(max) = sqrt((2(2.49 × 10^(-18) - 19.8×10^(-19)))/ 9.11×10^(-31))
⇒ v_(max) = 1.05 × 10^6 m s⁻¹