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3 years ago

‏Explain how an electron emitted by the photoelectric effect can have kinetic energy less than threshold energy ?

Physics

1 answer:

xenn [34]3 years ago

4 0

Answer:

the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

K = E - Ф

Explanation:

The photoelectric effect is the emission of electrons from the surface of a metal.

This was correctly explained by Einstein, in his explanation the energy of the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

E = hf

E = K + Ф

K = E - Ф

The energy of the photons is given by the Planck relation E = hf and according to Einstein the number of joints must be added

E = n hf

Therefore, depending on the value of this energy, the emitted electrons can have energy from zero onwards.

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

Svetradugi [14.3K]

Answer:

Total contraction on the Bar = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

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Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180 × 10³ N

Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

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A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to

nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket $V=10\ gallon$

Time taken to fill the bucket $t=50\ s$

so volume flow rate is $\dot{V}=\frac{10}{50}=0.2\ gal/s$

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$\dot{V}=0.0267\ ft^3/s$

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8 0

3 years ago