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Sergio039 [100]
2 years ago
12

Calculate the de broglie wavelength of a subatomic particle that is moving at 351 km/s if its mass is 9.11 à 10â31 kg. λ = hmuh

mu and h = 6.626 Ã 10â34 kgâm2/s
Chemistry
1 answer:
gregori [183]2 years ago
6 0

The de Broglie wavelength of a subatomic particle is 2.09 nm.

 λ = h m v = h

momentum : wherein 'h' is the Plank's steady. This equation pertaining to the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated the use of this relation is de Broglie wavelength.

Frequency is the ratio of velocity and wavelength in relation to hurry. In evaluation, wavelength refers back to the ratio of velocity and frequency.

Wavelength is the gap between the crests of waves or a person's fashionable mind-set. An instance of wavelength is the gap between the crest of two waves. An instance of wavelength is while you and some other character share the equal standard attitude and might for that reason speak properly.

calculation is given in the image below

de Broglie wavelength λ = h/mv

                                          = (6.626 * 10^-34)/9.1 * 10^-31 *351 *10^3

                                          = 2.07 *10^-9

                                 Hence, = 2.op nm    

Learn more about de Broglie wavelength here:-brainly.com/question/16595523

#SPJ4

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nexus9112 [7]

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by N_v.

It is depends on

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  • energy required for vacancy
  • Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
  • temperature (T).

N_v=Ne^{-\frac{Q_v}{kT} }

To find  equilibrium number of of vacancies we have find N.

N=\frac{N_A\ \rho}{A_{cu}}

Here ρ= 8.45 g/cm³  =8.45 ×10⁶m³

N_A= Avogadro Number = 6.023×10²³

A_{Cu}= 63.5 g/mole

N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}

   =8.01\times 10^{28 g/mole

Here Q_v=0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})

   =2.34×10²⁴ vacancies/ mole

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Answer:

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Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

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It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

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