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3 years ago

7

Violet light of wavelength 427 nm ejects electrons with a maximum kinetic energy of 0.684 eV from a certain metal. What is the w

ork function of this metal (in eV)?

1 answer:

frutty [35]3 years ago

5 0

Answer:

The work function of the metal is 2.226 eV.

Explanation:

Given;

wavelength of the violet light, λ = 427 nm = 427 x 10⁻⁹ m

maximum kinetic energy, K.E = 0.684 eV

The energy of the incident light is calculated as;

$E = hf = \frac{hc}{\lambda} = \frac{6.626 \ \times \ 10^{-34} \ \times\ 3\ \times \ 10^8 }{427 \ \times \ 10^{-9}} = 4.655 \ \times \ 10^{-19} \ J\\\\1 \ eV = 1.6 \ \times \ 10^{-19} \ J\\\\E =( \frac{4.655 \ \times \ 10^{-19} \ J }{1.6 \ \times \ 10^{-19} \ J} ) \ eV\\\\E = 2.91 \ eV$

Apply Einstein's photoelectric equation;

E = Ф + K.E

where;

Ф is the work function of the metal

Ф = E - K.E

Ф = 2.91 eV - 0.684 eV

Ф = 2.226 eV.

Therefore, the work function of the metal is 2.226 eV.

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3 years ago

A roller coaster car rapidly picks up speed as it rolls down a slope. As it starts down the slope, its speed is 4m/s. But 3 seco

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Answer:

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Explanation:

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where

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3 years ago

Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.

olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

$U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{x,1}$ , $K_{x,2}$ - Initial and final horizontal translational kinetic energy, measured in joules.

$K_{y,1}$ , $K_{y,2}$ - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

$m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})$

$y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}$

Where:

$y_{1}$ . $y_{2}$ - Initial and final height of the arrow, measured in meters.

$v_{y,1}$ , $v_{y,2}$ - Initial and final vertical speed of the arrow, measured in meters.

$g$ - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

$v_{y,1} = v_{1}\cdot \sin \theta$

Where:

$v_{1}$ - Magnitude of the initial velocity, measured in meters per second.

$\theta$ - Initial angle, measured in sexagesimal degrees.

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the initial vertical speed is:

$v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}$

$v_{y,1} \approx 33.352\,\frac{m}{s}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} \approx 33.352\,\frac{m}{s}$ and $v_{y,2} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{2} - y_{1} = 56.712\,m$

Second arrow

$U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}$

Where:

$U_{g,1}$ , $U_{g,3}$ - Initial and final gravitational potential energy, measured in joules.

$K_{y,1}$ , $K_{y,3}$ - Initial and final vertical translational kinetic energy, measured in joules.

$m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})$

$y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} = 82\,\frac{m}{s}$ and $v_{y,3} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{3} - y_{1} = 342.816\,m$

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

$E = U + K_{x}$

The expression is now expanded:

$E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}$

Where $v_{x}$ is the horizontal speed of the arrow, measured in meters per second.

$v_{x} = v_{1}\cdot \cos \theta$

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the horizontal speed is:

$v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}$

$v_{x} \approx 74.911\,\frac{m}{s}$

If $m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{max} = 56.712\,m$ and $v_{x} \approx 74.911\,\frac{m}{s}$ , the total mechanical energy is:

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}$

$E = 201.720\,J$

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

$E = m\cdot g \cdot y_{max}$

$m = 0.06\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $y_{max} = 342.816\,m$

$E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)$

$E = 201.720\,J$

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0

3 years ago

A cardinal (Richmondena cardinalis) of mass 4.20×10−2 kg and a baseball of mass 0.146 kg have the same kinetic energy. What is t

bekas [8.4K]

Explanation:

It is given that,

Mass of cardinal, $m_c=4.2\times 10^{-2}\ kg$

Mass of baseball, $m_b=0.146\ kg$

Both cardinal and baseball have same kinetic energy. We need to find the ratio of the cardinal's magnitude $p_c$ of momentum to the magnitude $p_c$ of the baseball's momentum.

$E_c=E_b$

Kinetic energy is given by, $E=\dfrac{p^2}{2m}$

$\dfrac{p_c^2}{2m_c}=\dfrac{p_b^2}{2m_b}$

$(\dfrac{p_c}{p_b})^2=\dfrac{m_c}{m_b}$

$\dfrac{p_c}{p_b}=\sqrt{\dfrac{m_c}{m_b}}$

$\dfrac{p_c}{p_b}=\sqrt{\dfrac{4.2\times 10^{-2}}{0.146}}$

So, $\dfrac{p_c}{p_b}=\dfrac{53}{100}$

So, the ratio of cardinal's magnitude of momentum to the magnitude of the baseball's momentum is 53 : 100

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3 years ago

A pendulum is timed, first for 20 swings and then for

wlad13 [49]

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2 years ago