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2 years ago

Nuclear forces exists because the particles in the nucleus are ?

Chemistry

1 answer:

zloy xaker [14]2 years ago

8 0

Explanation:

Nuclear forces exist because the particles in the nucleus are close together the tritium atom consists of one proton, two neutron, and one electron how many isotopes of hydrogen are known 3 The hydrogen isotope with the least mass is named protium

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Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

7 0

2 years ago

Why s orbital does not contain degenerate orbital

Juliette [100K]

There’s no other degenerate s orbital because the there can be only one s orbital for any value of n.

3 0

2 years ago

A gas sample is found to contain 39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine. If the mo

Doss [256]

Answer:

C6H14O3F

Explanation:

The first step is to divide each compound by its molecular weight

Carbon

= 39.10/12

= 3.258

Hydrogen

= 7.67/1

= 7.67

Oxygen

= 26.11/16

= 1.63

Phosphorous

= 16.82/31

= 0.542

Flourine

= 10.30/19

= 0.542

The next step is to divide by the lowes value

3.258/0.542

= 6 mol of C

7.67/0.542

= 14 mol of H

1.63/0.542

= 3 mol of O

0.542/0.542

= 1 mol of P

0.542/0.542

= 1 mol of F

Hence the molecular formula is C6H14O3F

5 0

2 years ago

A radioactive material, with half-life of six months, has 100 thousand unstable nuclei.

goldfiish [28.3K]

Answer:

See Explanation

Explanation:

Given that;

N/No = (1/2)^t/t1/2

Where;

No = amount of radioactive isotope originally present

N = A mount of radioactive isotope present at time t

t = time taken

t1/2 = half life

N/1000=(1/2)^3/6

N/1000=(1/2)^0.5

N = (1/2)^0.5 * 1000

N= 707 unstable nuclei

Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;

Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1

Hence;

A=Aoe^-kt

Where;

A = Activity after a time t

Ao = initial activity

k = decay constant

t = time taken

A = Aoe^-3 *0.1155

A=Aoe^-0.3465

3 0

2 years ago

When the sun shines and heats something, that energy is called-

vodomira [7]

Answer:

Solar energy.

5 0

2 years ago

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