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3 years ago

An electron has a kinetic energy of 3.00 ev. find its wavelength. (b) what if? a photon has energy 3.00 ev. find its wavelength.

1 answer:

8 0

(a) The electron kinetic energy is
$K=3.00 eV$
which can be converted into Joule by keeping in mind that
$1 eV=1.6 \cdot 10^{-19}eV$
So that we find
$K=3.00 eV \cdot 1.6 \cdot 10^{-19} eV/J =4.8 \cdot 10^{-19}J$

The kinetic energy of the electron is related to its momentum p by:
$K= \frac{p^2}{2m}$
where m is the electron mass. Re-arranging the equation, we find
$p= \sqrt{ 2Km}= \sqrt{ 2 ( 4.8 \cdot 10^{-19} J)(9.1 \cdot 10^{-31} kg) } =9.35 \cdot 10^{-25} kgm/s$

And now we can use De Broglie's relationship to find its wavelength:
$\lambda= \frac{h}{p}= \frac{6.6 \cdot 10^{-34} Js}{9.35 \cdot 10^{-25} kg m/s} =7.06 \cdot 10^{-10}m$
where h is the Planck constant.

(b) By using the same procedure of part (a), we can convert the photon energy into Joules:
$E=3.00 eV \cdot 1.6 \cdot 10^{-19} eV/J =4.8 \cdot 10^{-19}J$

The energy of a photon is related to its frequency f by:
$E=hf$
where h is the Planck constant. Re-arranging the equation, we find
$f= \frac{E}{h}= \frac{4.8 \cdot 10^{-19} J}{6.6 \cdot 10^{-34}Js} =7.27 \cdot 10^{14}Hz$

And now we can use the relationship between frequency f, speed of light c and wavelength $\lambda$ of a photon, to find its wavelength:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.27 \cdot 10^{14} Hz} =4.13 \cdot 10^{-7} m$

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A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s

Sonbull [250]

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

(I was only able to do A and B)

7 0

2 years ago

Read 2 more answers

Please answer ASAP, and please don't joke around and actually do answer my question. I will give you brainliest if you answer it

hammer [34]

Answer:

30 feet

Explanation:

First off, if they are throwing at 12.0 m/s and it takes 2.5 seconds. It will be the act of multiplication.

12 times 2.5 is 30, because 12 times 2 is 24 plus 12 divided by 2 which is 6 so 24 plus 6 is 30.

8 0

2 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$