Photoelectric Effect
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New answer posted
a month agoContributor-Level 10
hc/λ = K? + φ [given φ is negligible]
So, hc/λ = K?
λ? = h/√ (2mK? ²) ⇒ K? = h²/ (2mλ? ²)
(hc)/λ = h²/ (2mλ? ²) ⇒ λ = (2mc/h)λ? ²
New answer posted
a month agoContributor-Level 10
The equation representing the photoelectric effect is given as:
(1/2)mv? ² = hf? - φ . (1)
This continues the solution from number 6 on the previous page.
(1/2)mv? ² = hf? - φ . (2)
With the help of equations (1) and (2), we can write:
v? ² - v? ² = (2h/m) (f? - f? )
New answer posted
a month agoContributor-Level 10
eVs = hv – φ
At Vs = 0 ⇒ hv = φ
⇒ φ = [6.62 * 10?³?][10¹?][5.5]
⇒ φ = ([6.62 * 10?³?][10¹?][5.5]eV) / [1.6 * 10?¹?]
= 2.27
New answer posted
a month agoContributor-Level 10
If the work function of the metal is φ, then the kinetic energy (K.E.) of the emitted photoelectron is given by Einstein's photoelectric equation:
K.E. = hν - φ = (hc/λ) - φ
Case 1: λ? = 500 nm
K? = (hc/λ? ) - φ
Case 2: λ? = 200 nm
K? = (hc/λ? ) - φ
Given, K? = 3K?
So, (hc/λ? ) - φ = 3 * [ (hc/λ? ) - φ]
(hc/λ? ) - φ = 3 (hc/λ? ) - 3φ
2φ = 3 (hc/λ? ) - (hc/λ? )
2φ = hc * (3/λ? - 1/λ? )
φ = (hc/2) * [3/ (500 nm) - 1/ (200 nm)]
φ = (hc/2) * [ (6 - 5) / 1000 nm]
φ = hc / 2000 nm
Using hc ≈ 1240 eV·nm
φ = 1240 eV·nm / 2000 nm = 0.62 eV
New answer posted
a month agoContributor-Level 10
E? = φ + K?
E? = φ + K?
E? - E? = K? - K?
Now V? /V? = 2
K? /K? = 4; K? = 4K?
Now from equation (2)
⇒ 4 - 2.5 = 4K? - K?
1.5 = 3K?
K? = 0.5eV
Now putting this
Value in equation (2)
2.5 = φ + 0.5eV
φ = 2eV
New answer posted
2 months agoContributor-Level 10
Stopping potential defined in terms of wavelength as :
Using above two equation we can calculate :
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