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3 years ago

6

If we have a one-electron atom, and the electron can occupy any of 5 different energy levels, and transitions are possible betwe

en any level and any other, how many lines might appear in that atom’s spectrum?

1 answer:

xeze [42]3 years ago

7 0

Answer:

10

Explanation:

Lets say that a,b,c,d,e are the five allowed energy states in order of decreasing energy. Then the number of possible different spectral lines comes from the electron dropping from a high state to a lower state. So, they will do so in following ways:

a - b

a - c

a - d

a- e

b - c

b - d

b- e

c - d

c- e

d- e

Ans- Ten different possible energy jumps giving six different colors or lines in the spectrum.

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1 lb equals how many grams

shusha [124]

Answer:

1 lb. is 453.592 grams

Explanation:

1lb is 453.592 grams

8 0

3 years ago

Read 2 more answers

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

6 0

3 years ago

How long will it take you to travel 20 miles on a bus that drives 60 miles/h?

ValentinkaMS [17]

Answer:

<u>20 Minutes</u>

<u></u>

Explanation:

Well we know Mph (Miles per hour) is distance over time : $\frac{distance}{time} \\$

R (rate) = 60

d (distance) = 20

t (time) = Unknown

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

R = $\frac{d}{t}$

↓

60 = $\frac{20}{t}$

↓

t = $\frac{20}{60}$

↓

t = $\frac{1}{3}$ or 0.3333

<em>So basically it would take one third of an hour. Lets change these units to minutes.</em>

60 * 0.333333 = 20

<em>So it would take you </em><u><em>20 minutes</em></u><em> to drive 20 miles on a bus that drives 60 mph</em>

<em />

Hope that helps

<em>~Siascon~</em>

8 0

3 years ago

What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c

Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

= 413V / 0.0795 m

= 5194.97 V/M

Explanation:

the potential difference between two uniform plates is calculated by the formula of electric field.

6 0

3 years ago

Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes tha

Dahasolnce [82]

Answer:

Sarah is right

Explanation:

This is an exercise that differentiates between scalars and vectors.

A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.

In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero

consequently Sarah is right

4 0

3 years ago