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Lapatulllka [165]
2 years ago
9

Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation21484 Po -->

21082 Pb + 42 HeAtomic masses: Pb-210 = 209.98284 amu,Po-214 = 213.99519 amu, He-4 = 4.00260 amu. (1 kg = 6.022 × 1026 amu; NA = 6.022 × 1023 mol–1; c = 2.99792458 × 108 m/s)
Physics
1 answer:
GuDViN [60]2 years ago
8 0

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

Atomic masses: Pb-210 = 209.98284 amu

Po-214 = 213.99519 amu

He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

energy of molecule using equation

ΔE = Δm c²

Δm is mass difference and c is speed of light

Δm = 209.98284 + 4.00260 - 213.99519

Δm = - 0.00975 amu

1 amu = 1.66 x 10⁻²⁷ kg

- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

                         = -0.016185 x 10⁻²⁷ Kg

total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

                 = -0.097467 x 10⁻⁴ Kg

ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²

ΔE = - 8.77 × 10¹¹

ΔE = 8.77 × 10¹¹ J

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Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

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In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

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