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2 years ago

Siven an element's atomic number and mass number, how can you tell the number of protons and neutrons in its nucleus?

Physics

2 answers:

Ivahew [28]2 years ago

7 0

Answer:

Number of protons = atomic number; number of neutrons = mass number - atomic number

Explanation:

Z = atomic number = number of protons

A = mass number = number of protons + number of neutrons

Mass number = atomic number + number of neutrons

Number of neutrons = mass number - atomic number

sukhopar [10]2 years ago

3 0

It would be the first option.

Explanation-
The number of protons is equal to the atomic number the number of neutrons is the mass minus the atomic number.

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Could someone tell me if this is right or give me the answers so i can re write them! That would be so helpful:)

Ivahew [28]

Answer:

Those are all right

Explanation:

6 0

2 years ago

Read 2 more answers

The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

Finally:

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

2 years ago

A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate

slava [35]

The refractive index of water is $n=1.33$ . This means that the speed of the light in the water is:
$v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s$

The relationship between frequency f and wavelength $\lambda$ of a wave is given by:
$\lambda= \frac{v}{f}$
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water $\lambda_w$ to that in air $\lambda$ as
$\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}$
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using $\lambda=400 nm$ , we find
$\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm$
which is the wavelength of light in water.

5 0

2 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$