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2 years ago

How does the current model of the atom differ from j j thomas model

Physics

1 answer:

bixtya [17]2 years ago

4 0

Jj thomas' model contained electrons.

The current model contains a nucleus with electrons orbiting around it. :)

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The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc

Ivanshal [37]

1. Frequency: $3.23\cdot 10^{20} Hz$

The energy given is the energy per mole of particles:

$E=1.29\cdot 10^{11} J/mol$

1 mole contains a number of Avogadro of particles, $N_A$ , equal to

$N_A=6.022\cdot 10^{23}$ particles

So, by setting the following proportion, we can calculate the energy of a single photon:

$1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J$

This is the energy of a single photon; now we can calculate its frequency by using the formula:

$E_1 = hf$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck's constant

f is the photon frequency

Solving for f, we find

$f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz$

2. Wavelength: $9.29\cdot 10^{-13} m$

The wavelength of the photon is given by the equation:

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

where

$c=3\cdot 10^8 m/s$

is the speed of the photon (the speed of light). Substituting,

$\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m$

6 0

2 years ago

A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t

Jobisdone [24]

Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

E= 411.84 * 10³ N/C

4 0

3 years ago

an object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m. determine the image height (in cm)

Juli2301 [7.4K]

Given :

An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

To Find :

the image height (in cm).

Solution :

By lens formula :

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

Here, u = - 100 cm

f = 150 cm

$\dfrac{1}{v} - \dfrac{1}{-100} = \dfrac{1}{150}\\\\v = - 300 \ cm$

Now, magnification is given by :

$m = \dfrac{v}{u} = \dfrac{h_i}{h_o}\\\\h_i = \dfrac{300}{100}\times 1\\\\h_i = 3 \ m$

Therefore, the image height is 3 m or 300 cm.

5 0

2 years ago

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =

Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = $0.5\times 10^{-4} kg$

Weight of the rain drop = W

Duration of time = t = 3 seconds

$W=m\times g$

Drag force on rain drop = $D=0.2\times 10^{-5} v^2$

$W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N$

Motion of the rain drop:

$F=m\times a$

Net force on the rain drop , F= W - D

$W-D=m\times a$

$0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a$

$0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}$

$0.006v^2+0.05v-1.5=0$

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

$12.18 m/s=0m/s+a\times 3 s$

$a=4.06 m/s^2$

$s=ut+\frac{1}{2}at^2$ (second equation of motion)

$s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2$

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0

2 years ago

A cannon can make a maximum angle of 30 degrees with the horizon. What is the minimum speed of a cannon ball if it must clear a

stiks02 [169]

Answer:

28.3 m/s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 30°

Maximum height (H) = 10 m

Acceleration due to gravity (g) = 10 m/s²

Initial velocity (u) =?

Thus, we can obtain the minimum velocity cannon ball by using the following formula:

H = u²Sine² θ / 2g

10 = u² × (Sine 30)² / 2× 10

10 = u² × (0.5)² / 20

10 = u² × 0.25 / 20

10 = u² × 0.0125

Divide both side by 0.0125

u² = 10/ 0.0125

u² = 800

Take the square root of both side

u = √800

u = 28.3 m/s

Therefore, the minimum speed of the cannon ball is 28.3 m/s

8 0

2 years ago