What is it? Please help me.

When 224-nm light falls on a metal, the current through a photoelectric circuit is brought to zero at a stopping voltage of 1.84

OLga [1]

Answer:

3.71 eV

Explanation:

λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m

c = speed of electromagnetic wave = 3 x 10⁸ m/s

V₀ = stopping potential = 1.84 volts

W₀ = Work function of the metal = ?

Using the equation

$\frac{hc}{\lambda } = eV_{o} + W_{o}$

$\frac{(6.63\times 10^{-34})(3\times 10^{8})}{224\times 10^{-9} } = (1.6\times 10^{-19})(1.84) + W_{o}$

$W_{o}$ = 5.94 x 10⁻¹⁹

$W_{o}$ = 3.71 eV

7 0

2 years ago

What is one factor that helps keep the planets and other solar system objects in orbit around the sun?

pshichka [43]

The answer is D<span>.the gravitational force between the sun and each object in the solar system</span>

7 0

3 years ago

Read 2 more answers

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectivel

faust18 [17]

Answer:

The time is 5.10 sec.

Explanation:

Given that,

Component of velocity are 30 m/s and 40 m/s.

We need to calculate the resultant velocity

Using formula of resultant velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(30)^2+(40)^2}$

$v=50\ m/s$ [

We need to calculate the time

Using equation of motion

$v = u+gt$

$v = 0+gt$

$t=\dfrac{v}{g}$

Put the value into the formula

$t=\dfrac{50}{9.8}$

$t=5.10\ sec$

Hence, The time is 5.10 sec.

7 0

2 years ago

Copernicus incorrectly assumed that the planets?

hammer [34]

Nicholas Copernicus correctly assumed that the planets revolved around the sun but he incorrectly assumed that the planets followed a perfect circle orbit around the sun. It was later on discovered by Johannes Kepler that the planets moved around the sun following an elliptical orbit.

6 0

2 years ago

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a

BlackZzzverrR [31]

Answer:

E = 8.5 * 10^6 V/m

Explanation:

In general we have the following relation between the Electric Field and the Elecric Potential:

$\int\limits^2_1 {\vec{E} \cdot\vec{dl}} = V_{2} -V_{1}$

Due to the vector nature of the electric filed, we can only know the mean Electric field E across the membrane, and take it out from the integral, that is:

E = (ΔV)/L

Where L is the thickness of the membrane and ΔV is the potential difference.

Therefore:

E = 8.53933*10^6 V/m

rounding to the first tenth:

E = 8.5 * 10^6 V/m

3 0

2 years ago