Ask question

Ask question

All categories

3 years ago

11

When light shines through atomic hydrogen gas, it is seen that the gas absorbs light readily at a wavelength of 91.63 nm. What i

s the value of the principal quantum number n of the level to which the hydrogen is being excited by the absorption of light of this wavelength? Assume that the most of the atoms in the gas are in the lowest level. (A) 14 (B) 16 (C) 11 (D) 21

1 answer:

Artist 52 [7]3 years ago

4 0

Answer:

D) 21

Explanation:

When gas absorbs light , electron at lower level jumps to higher level .

and the difference of energy of orbital is equal to energy of radiation absorbed.

Here energy absorbed is equivalent to wavelength of 91.63 nm

In terms of its energy in eV , its energy content is eual to

1243.5 / 91.63 = 13.57 eV. This represents the difference the energy of orbit .

Electron is lying in lowest or first level ie n = 1.

Energy of first level

= - 13.6 / 1² = - 13.6 eV.

Energy of n th level = - 13.6 / n². Let in this level electron has been excited

Difference of energy

= 13.6 - 13.6 / n² = 13.57 ( energy of absorbed radiation)

13.6 / n² = 13.6 - 13.57 = .03

n² = 13.6 / .03 = 453

n = 21 ( approx )

You might be interested in

Explain why the moon is always half illuminated and half dark no matter where it is in the lunar cycle.

BlackZzzverrR [31]

The easiest way I know to explain it is this:

-- Take a flashlight and a ball into a dark room.

-- Turn on the flashlight and point it at the ball.

-- Half of the ball is lighted up by the flashlight, and the other half is dark.

-- There is no way you can turn or twist the ball to make more or less
than 50% of it lighted up and more or less than 50% of it dark.

<em>Everything</em> in the solar system ... as long as it's shaped like a ball ... is
half illuminated by the sun and half dark.

7 0

2 years ago

What substance can be used to electrolyze water?

lubasha [3.4K]

C/any electrolyte that is not easily reduced or oxidized

6 0

3 years ago

Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground

Papessa [141]

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that $sin(betha)=(leg opposite)/(hypotenuse)$

In this case we will need to find β, this way:

$betha=sin^-1((15ft/20ft))$

Then β=48,6°

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

$betha=(48,6°)*(2pirads/360°)$

then we find β=0,85rads

7 0

2 years ago

What information do you think the temperatures of stars give us?

Paul [167]

Explanation:

One way of classifying stars is by their temperature .

or

Science strives to be able to describe how stars and planets form and evolve. This requires theories to describe the processes which include:

Star and planet formation

Star and planet composition

Stellar and solar system evolution

The nuclear processes happening inside stars

The scientific method means that all theories are put to the test. By measuring or calculating the temperature, age and composition of other planets and stars the theories can be tested. If observed values of these parameters are not predicted by theories, then the theories are wrong and need to be revised or replaced.

6 0

2 years ago

Read 2 more answers

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

So

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago