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1 year ago

Describe how electromagnetic radiation can ionise an atom. 2 marks

Physics

1 answer:

IRISSAK [1]1 year ago

8 0

Answer:

Ionizing radiation is radiation with enough energy so that during an interaction with an atom, it can remove tightly bound electrons from the orbit of an atom, causing the atom to become charged or ionized. ... Forms of electromagnetic radiation.

(from google)

thank you :)

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A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n

Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

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

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

$\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m$

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

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

where for the dolphin:

v = 1531 m/s is the wave speed

$\lambda=1.50 cm=0.015 m$ is the wavelength

Substituting into the equation,

$f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz$

C. 13.6 m

Again, the wavelength is given by:

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

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

$\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m$

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

$\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m$

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

$\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m$

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m$

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$

3 0

2 years ago

There are lots of examples of ideal gases in the universe, and they exist in many different conditions. In this problem we will

elena-14-01-66 [18.8K]

Answer:

P = ρRT/M

Explanation:

Ideal gas equation is given as follows generally:

PV = nRT (1)

P = pressure in the containing vessel

V = volume of the containing vessel

n = number of moles

R = gas constant

T = temperature in K

n = m/M

m = mass of the gas contained in the vessel in g

M = molar mass in g/mol

ρ = m/V

Density of the gas = ρ

Substituting for n in (1)

PV = mRT/M. (2)

Dividing equation (2) through by V

P = m/V ×RT/M

P = ρRT/M

5 0

1 year ago

A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by

Airida [17]

Answer:

Mobility of the minority carriers, $\mu_{n} =1184.21 cm^{2} /V-sec$

Diffusion coefficient for minority carriers, $D_{n} = 29.20 cm^2 /s$

Verified from Einstein relation as $\frac{D_{n} }{\mu_{n} } = 25 mV$

Explanation:

Length of sample, $l_{s} = 2 cm$

Separation between the two probes, L = 1.8 cm

Drift time, $t_{d} = 0.608 ms$

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), $\mu_{n} = \frac{V_{d} }{E}$

Where the drift velocity, $V_{d} = \frac{L}{t_{d} }$

$V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s$

and the Electric field strength, $E = \frac{V}{l_{s} }$

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

$\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec$

The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

$\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV$

Verified.

4 0

2 years ago

A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag

klio [65]

Answer:

57.94°

Explanation:

we know that the expression of flux

$\Phi =E\times S\times COS\Theta$

where Ф= flux

E= electric field

S= surface area

θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 $\frac{Nm^{2}}{sec}$

E= $1.44\times 10^{4}\frac{Nm}{C}$

S= $\pi \times 0.057^{2}$

$COS\Theta =\frac{\Phi }{S\times E}$

= $\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}$

=0.5306

θ=57.94°

4 0

2 years ago

What is the formula that can be used to find velocity if kinetic energy and mass are known?

viva [34]

The formula for kinetic energy is $\frac{1}{2}m\Delta v^2$ . Thus, the equation for velocity is $v= \sqrt{ \frac{2TotalKineticEnergy}{m} }$ .

6 0

2 years ago

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