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3 years ago

Which colors will eject electrons when they strike sodium, which has a frequency threshold of 5.7 × 1014 Hz?

Physics

2 answers:

Harlamova29_29 [7]3 years ago

8 0

Answer: Violet and Blue color

Explanation:

From photoelectric effect, we understand that when a radiation of frequency equal or greater than threshold frequency of metal falls on it, electrons are ejected. Thus, the color which will eject electrons when they strike sodium which has a threshold frequency of 5.7 × 10¹⁴ Hz should have greater frequency than its.

The frequency of Blue color is 6.6× 10¹⁴ Hz and violet is 7.3 × 10¹⁴ Hz . Thus, these two colors would be able to eject electrons from Sodium. Rest of the colors have lower frequency than threshold.

sineoko [7]3 years ago

6 0

Violet and green colour

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Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

3 0

3 years ago

A sprinter accelerates at 7.5 m/s from rest in 2.0 s, what distance did she go? (15 m)

a_sh-v [17]

Answer:

The sprinter traveled a distance of 7.5 m

Explanation:

Motion With Constant Acceleration

It's a type of motion in which the rate of change of the velocity of an object is constant.

The equation that rules the change of velocities is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:

$\displaystyle a=\frac{7.5-0}{2}$

$a=3.75\ m/s^2$

Now calculate the distance:

$\displaystyle x=0*2+\frac{3.75*2^2}{2}$

$\displaystyle x=7.5\ m$

The sprinter traveled a distance of 7.5 m

8 0

3 years ago

Explain what a coulomb is and be able to change from electron to coulomb and vice versa

torisob [31]

Answer:

Cuolomb is SI unit of charge

1 e=1.6022E-19 C

1 C = 6.2415E+18 e

Explanation:

Coulomb(C) is the SI unit for measuring electric charge and is given as the product of current and time. To convert Cuolomb to electron, we multiply each cuolomb of charge by 1 C = 6.2415E+18 e while to convert electron into charge, we multiply the electrons by 1e=1.6022E-19 C

4 0

3 years ago

How did you know if an electric current is flowing in a lightbulb?

sweet [91]

Answer:

if it lights up

Explanation:

electricity

7 0

3 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

4 years ago