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2 years ago

in the bohr model, which energy transition will result in light being emitted with shortest wavelength?

Physics

1 answer:

RoseWind [281]2 years ago

7 0

The line of shortest wavelength is from n4 - n1.

<h3>What is bohr model ?</h3>

The Bohr model, also known as the Rutherford-Bohr model, was proposed by Niels Bohr and Ernest Rutherford in 1913 and depicts an atomic system with an orbiting system of electrons surrounding a small, dense nucleus. This arrangement is similar to that of the Solar System, except that electrostatic forces act as the forces of attraction rather than gravity.

The line with the shortest wavelength in the hydrogen spectrum is created when an electron in an atom of hydrogen transitions between the first and fourth energy levels (n4 →n1).

The transition n4→ n1 has the largest energy difference between the transitions that are given.

Lower wavelengths correspond to higher energy differences.

to learn more about bohr's model go to -

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A light spring of constant 163 N/m rests vertically on the bottom of a large beaker of water.

LUCKY_DIMON [66]

Answer:

24.71cm

Explanation:

We approach this problem base don Hooke's law which states the elongation produced in an elastic material is proportional to the applied load or force provided that its elastic limit is not exceeded. This is expressed mathematically as follows;

$F=ke................(1)$

where F is the applied force, k is the force constant and e is the elongation or extension of the material.

In this problem, the applied force F is the weight of the wood which is calculated as follows,

$F=mg.............(2)$

m = 4.11kg

$g=9.8m/s^2$

Hence,

$F=4.11*9.8\\F=40.278N$

Given that k = 163N/m, we make appropriate substitutions into equation (1) to obtain the following;

$40.278=163*e\\e=\frac{40.78}{163}\\e=0.2471m$

Since it is required in cm, we perform the conversion as follows, knowing that 100cm = 1m

$0.2471m=0.2471*100cm=24.71cm$

NB: We do not necessarily need the the density of the wood to perform our calculations since other parameters were given from which we were able to obtain its weight.

7 0

4 years ago

Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

Romashka [77]

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

$x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}$

Camioneta lenta

$x_{C} = x_{C,o} +v_{C}\cdot t$

Donde:

$x_{S,o}$ , $x_{C,o}$ - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

$v_{S,o}$ - Velocidad inicial de Speedy Sue, medida en metros por segundo.

$v_{S}, v_{C}$ - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

$a_{S}$ - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

$t$ - Tiempo, medido en segundos.

Si conocemos que $x_{S} = x_{C}$ , $x_{S,o} = 0\,m$ , $x_{C,o} = 155\,m$ , $v_{S,o} = 30\,\frac{m}{s}$ , $v_{C} =-5\,\frac{m}{s}$ y $a_{S} = -2\,\frac{m}{s^{2}}$ , encontramos la siguiente función cuadrática:

$155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}$

$-t^{2}+35\cdot t-155 = 0$ (Ec. 1)

Las raíces de esta función son:

$t_{1}\approx 29.798\,s$ , $t_{2} \approx 5.201\,s$

La colisión ocurriría en la raíz positiva más pequeña, es decir:

$t \approx 5.201\,s$

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: ( $v_{C,o} = -5\,\frac{m}{s}$ , $x_{C,o} = 155\,m$ , $t \approx 5.201\,s$ )