Dual Nature of Radiation and Matter
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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
hc/λ = Φ + eV
hc/3λ = Φ + eV/4
from (1) and (2)
hc/λ (1-1/3) = 3/4 eV
eV = 8/9 hc/λ
eV = 8/9 (hc/λ - Φ)
Φ = hc/9λ
λ? = 9λ ∴k=9
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
a month agoContributor-Level 10
Using de-Broglie equation:
λ = h/p = h/√ (2mK) = h/√ (2meV)
λ ∝ 1/√m
λ? /λ? = √ (m? /m? ) = √ (1.00727/0.00055) = √1831 ≈ 43:1
New answer posted
a month agoContributor-Level 10
K = hc/λ - W?
K = (20*10? ²? )/ (500*10? ) - 1.25 * 1.6*10? ¹?
K = 4*10? ¹? - 2*10? ¹? = 2*10? ¹? J
R = √ (2mK)/ (eB)
B = √ (2mK)/ (eR) = √ (2 * 9*10? ³¹ * 2*10? ¹? ) / (1.6*10? ¹? * 30*10? ²)
B = √ (36*10? ) / (48*10? ²¹) = (6*10? ²? ) / (4.8*10? ²? ) = 1.25 * 10? T
B = 125 * 10? T
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