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3 years ago

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Ultraviolet light of wavelength 3500 Angstrom falls on a potassium surface. The maximum energy of the emitted photoelectrons is

1.6 eV. Find the work function of potassium.

1 answer:

slega [8]3 years ago

8 0

Answer:

The work function of potassium is 1.94 eV.

Explanation:

Given that,

Wavelength $\lambda= 3500\ \AA$

Kinetic energy = 1.6 eV

We need to calculate the energy

Using formula of Energy

$E=\dfrac{hc}{\lambda}$

Put the value into the formula

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3500\times10^{-10}}$

$E=5.68\times10^{-19}\ J$

$E=5.68\times10^{-19}\times6.24\times10^{18}\ ev$

$E=3.54\ ev$

We need to calculate the work function of potassium

Using formula of work function

$E_{x}=\phi+k_{max}$

Put the value into the formula

$3.54=\phi+1.6$

$\phi=3.54-1.6$

$\phi=1.94\ ev$

Hence, The work function of potassium is 1.94 eV.

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

4 0

3 years ago

What causes a compass needle to point to geographic north?

morpeh [17]

I believe it’s “C”
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3 years ago

Why is the water cycle important to life on Earth?

sertanlavr [38]

C. The water cycle spreads water out evenly around the whole Earth

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4 years ago

Read 2 more answers

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

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3 years ago

Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a frequency of 3.0 x 1012 H

djyliett [7]

Answer:

The wavelength of the infrared wave is <u>0.0001 m</u>.

Explanation:

Given:

Frequency of an infrared wave is, $f=3.0\times 10^{12}\ Hz$

We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.

So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:

$v=c=3.0\times 10^8\ m/s$

Where, 'v' is the speed of infrared waves and 'c' is the speed of light.

Now, we have a formula for the speed of any wave and is given as:

$v=f\lambda$

Where, $\lambda \to \textrm{Wavelength of infrared wave}$

Now, rewriting the above formula in terms of wavelength, $\lambda$ , we get:

$\lambda=\dfrac{v}{f}$

Now, plug in $3.0\times 10^8$ for 'v', $3.0\times 10^{12}$ for 'f' and solve for $\lambda$ . This gives,

$\lambda=\frac{3.0\times 10^8}{3.0\times 10^{12}}\\\\\lambda=0.0001\ m$

Therefore, the wavelength of the infrared wave is 0.0001 m.

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3 years ago