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3 years ago

6

In a particular television picture tube, the measured beam current is 23.3 µA . How many electrons strike the tube screen every

28 s ? The fundamental charge is 1.602 × 10−19 C. Answer in units of electrons.

1 answer:

zysi [14]3 years ago

7 0

Answer:

Explanation:

Given that

Beam current (i)=23.3µA

And the time to strike(t)=28s

Also, a fundamental charge e=1.602×10^-19C

Then, the charge quantity is given as,

q=it

Then, q=23.3×10^-6×28

q=6.524×10^-4C

Also, the number of electron N is given as

q=Ne

Therefore, N=q/e

So, N=6.524×10-4/1.602×10^-19

N=4.072×10^15

There are 4.072×10^15 electrons strike the tube screen every 28 s.

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Answer: 9.9%

Explanation: efficiency = (work output /work input) × 100

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Explanation:

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2 years ago

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

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Answer:

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Explanation:

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The second order of dark fringe considered is $m_2 = 6$

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$y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=> $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

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3 years ago

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3 years ago

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

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<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

3 years ago