Physics, 27.11.2019 04:31 R3nNyx

White light, with frequencies ranging from 4.00 x 10^14 hz to 7.90 x 10^14 hz, is incident on a barium surface. given that the work function of barium is 2.52 ev, find:
(a) the maximum kinetic energy of electrons ejected from this surface.
(b) the range of frequencies for which no electrons are ejected.

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Answer from: looloo273

0.7515875 eV

$4\times 10^{14}\leq f$

Explanation:

f = Maximum frequency = $7.9\times 10^{14}\ Hz$

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

W = Work function = 2.52 eV

Converting to Joules

$W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J$

Maximum photon energy is given by

$E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J$

Maximum Kinetic energy is given by

$K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J$

Converting to eV

$1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV$

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

$W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz$

The range of frequencies for which no electrons are ejected is

$4\times 10^{14}\leq f$

Answer from: Quest

ι вєℓιєνє ιт ιѕ тнє fιяѕт ¢нσι¢є(α.)

нє ѕнσυℓ∂ ѕмιℓє ρσℓιтℓу αи∂ ѕρєαк тσ тнє ρєяѕσи ωнσ ιѕ ιитєяяυριиg αfтєя тнє ∂ιѕ¢υѕѕισи

нσρє ι ¢συℓ∂ нєℓρ.

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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