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Volgvan
3 years ago
8

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2. Compute the wavelength of this line for l H a

nd 21. Note: These are the electronic hydrogen and deuterium atoms, not the muonic forms.]
Physics
1 answer:
irina1246 [14]3 years ago
6 0

Answer:

\lambda=550\ nm

Explanation:

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. n_i=3 and n_f=2

The wavelength of Hi line of the Balmer series is given by :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f}-\dfrac{1}{n_i})

\dfrac{1}{\lambda}=1.09\times 10^7\times (\dfrac{1}{2}-\dfrac{1}{3})

\lambda=5.50\times 10^{-7}\ m

\lambda=550\ nm

So, the wavelength for this line is 550 nm. Hence, this is the required solution.

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A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.36
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Answer: the average speed of the rat from the information given above is 0.7m/s

Explanation:

position is given as

x(t) = pt² + qt

finding the diffencial of x(t) with respect to t, we have

d(x(t))/dt = 2pt + q

we substitute the p = 0.36m/s² and q= -1.10 m/s

d(x(t))/dt = 2(0.36)t + (-1.10)

so, at t= 1s

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5 0
3 years ago
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

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