Question

What would the speed of each particle be if it had the same wavelength as a photon of yellow light (????=575.0 nm)? Proton (mass=1.673 x 10^−24 g) Neutron (mass = 1.675 x 10–24 g)Electron (mass = 9.109 x 10^–28 g)Alpha particle (mass = 6.645 x 10^–24 g)

Answer 1

Speed of Proton is about 0.6892 m/s

Speed of Neutron is about 0.6884 m/s

Speed of Electron is about 1266 m/s

Speed of Alpha particle is about 0.1735 m/s

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Further explanation

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

$\large {\boxed {E = h \times f}}$

where:

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

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Let's recall De Broglie's Wavelength Formula as follows:

$\boxed{\lambda = \frac{h}{mv}}$

where:

λ = wavelength ( m )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

m = mass of object ( kg )

v = velocity of object ( m/s )

Let us now tackle the problem !

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Given:

wavelength of yellow light = λ = 575.0 nm = 5.75 × 10⁻⁷ m

mass of proton = m₁ = 1.673 × 10⁻²⁴ g = 1.673 × 10⁻²⁷ kg

mass of neutron = m₂ = 1.675 × 10⁻²⁴ g = 1.675 × 10⁻²⁷ kg

mass of electron = m₃ = 9.109 × 10⁻²⁸ g = 9.109 × 10⁻³¹ kg

mass of alpha particle = m₄ = 6.645 × 10⁻²⁴ g = 6.645 × 10⁻²⁷ kg

Asked:

speed of each particle = v = ?

Solution:

We will use the formula of The Broglie's Wavelength:

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

$\boxed{v = \frac{h}{m \lambda}}$

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Proton:

$v_1 = \frac{h}{m_1 \lambda}$

$v_1 = \frac{6.63 \times 10^{-34}}{1.673 \times 10^{-27} \times 5.75 \times 10^{-7}}$

$v_1 \approx 0.6892 \texttt{ m/s}$

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Neutron:

$v_2 = \frac{h}{m_2 \lambda}$

$v_2 = \frac{6.63 \times 10^{-34}}{1.675 \times 10^{-27} \times 5.75 \times 10^{-7}}$

$v_2 \approx 0.6884 \texttt{ m/s}$

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Electron:

$v_3 = \frac{h}{m_3 \lambda}$

$v_3 = \frac{6.63 \times 10^{-34}}{9.109 \times 10^{-31} \times 5.75 \times 10^{-7}}$

$v_3 \approx 1266 \texttt{ m/s}$

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Alpha particle:

$v_4 = \frac{h}{m_4 \lambda}$

$v_4 = \frac{6.63 \times 10^{-34}}{6.645 \times 10^{-27} \times 5.75 \times 10^{-7}}$

$v_4 \approx 0.1735 \texttt{ m/s}$

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Learn more

• Photoelectric Effect : brainly.com/question/1408276
• Statements about the Photoelectric Effect : brainly.com/question/9260704
• Rutherford model and Photoelecric Effect : brainly.com/question/1458544

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Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics

Answer 2

Answer:

Proton: v=0.689 m/s

Neutron: v=0.688 m/s

Electron: v=1265.078 m/s

Alpha particle: v=0.173 m/s

Explanation:

De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:

λ=$\frac{h}{mv}$

h is the Planck constant: 6.626×10⁻³⁴$\frac{kg.m^2}{s}$

We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:

λ=$\frac{h}{mv}$

v=h÷(mλ)

Proton:

m=1.673×10⁻²⁴ g · $\frac{1kg}{1000g}$=1.673×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴$\frac{kg.m^2}{s}$÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)

v=0.689 m/s

Neutron:

m=1.675×10⁻²⁴ g · $\frac{1kg}{1000g}$=1.675×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴$\frac{kg.m^2}{s}$÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)

v=0.688 m/s

Electron:

m= 9.109×10⁻²⁸ g · $\frac{1kg}{1000g}$=9.109×10⁻³¹ kg

v=h÷(mλ)

v=6.626×10⁻³⁴$\frac{kg.m^2}{s}$÷(9.109×10⁻³¹ kg×575×10⁻⁹m)

v=1265.078 m/s

Alpha particle:

m=6.645×10⁻²⁴ g · $\frac{1kg}{1000g}$=6.645×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴$\frac{kg.m^2}{s}$÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)

v=0.173 m/s