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3 years ago

7

A ball is thrown with an initial speed of 20m/s at an angle of 60 to the ground. If air resistance is negligible, what is the ba

ll ‘s speed at the instant it reaches its maximum height from the ground.

1 answer:

torisob [31]3 years ago

4 0

Answer:

when a ball is thrown to the air and it attains it maximum height it's value is 0.

Explanation:

Therefore

V = Ucos angle

V = 20 cos 60°

V = 20 x0. 5

V= 10m/s

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An electron is trapped in an infinite square-well potential of width 0.6 nm. If the electron is initially in the n = 4 state, wh

grandymaker [24]

Answer:

E₁ = 1.042 eV

E₄₋₃= 7.29 eV

E₄₋₂= 12.50 eV

E₄₋₁= 15.63 eV

E₃₋₂= 5.21eV

E₃₋₁= 8.34eV

E₂₋₁= 3.13eV

Explanation:

The energy in an infinite square-well potential is giving by:

$E = \frac {h^{2} n^{2}}{8mL^{2}}$

<em>where, h: Planck constant = 6.62x10⁻³⁴J.s, n: is the energy state, m: mass of the electron and L: widht of the square-well potential </em>

<u>The energy of the electron in the ground state, </u><u>n = 1</u><u>, is: </u>

$E_{1} = \frac {(6.62 \cdot 10^{-34})^{2} (1)^{2}}{(8) (9.11 \cdot 10^{-31}) (0.6 \cdot 10^{-9} m)^{2}}$

$E_{1} = 1.67 \cdot 10^{-19} J = 1.042 eV$

The photon energies that are emitted as the electron jumps to the ground state is the difference between the states:

$E_{\Delta n} = \Delta n^{2} E_{1}$

$E_{(4 - 3)} = (4^{2} - 3^{2}) 1.042 eV = 7.29eV$

$E_{(4 - 2)} = (4^{2} - 2^{2}) 1.042 eV = 12.50eV$

$E_{(4 - 1)} = (4^{2} - 1^{2}) 1.042 eV = 15.63eV$

$E_{(3 - 2)} = (3^{2} - 2^{2}) 1.042 eV = 5.21eV$

$E_{(3 - 1)} = (3^{2} - 1^{2}) 1.042 eV = 8.34eV$

$E_{(2 - 1)} = (2^{2} - 1^{2}) 1.042 eV = 3.13eV$

Have a nice day!

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