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SVETLANKA909090 [29]
3 years ago
13

An electron is confined to a one dimensional infinite potential well 150 pm. How much energy must it absorb if it is to jump to

the state n-3 from ground state?
Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

\Delta E=1.22\times 10^{-22}J

Explanation:

The energy of electron in any state is given by E=\frac{n^2h^2}{8mL^2} here h is planck's constant n is state of electron L is the infinte potential well m is the mass of electron

We know that h=6.67\times 10^{-34}

Potential well dimension = 150pm=150\times 10^{-12}m

Mass of electron =9.1\times 10^{-31}kg

So energy required to electron to jump from ground state to 3rd state

\Delta E=\frac{h^2}{8mL^2}\left ( 3^2-1^2 \right )

\Delta E=\frac{\left ( 6.67\times 10^{-34} \right )^2}{8\times 9.1\times 10^{-31}(150\times 10^{-12})^2}\left ( 9-1 \right )

\Delta E=1.22\times 10^{-22}J

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11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

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5 0
3 years ago
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

6 0
3 years ago
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erastova [34]

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