The most common listening problem is

The Balmer series in hydrogen atom produces only infrared emission. O True O False

lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

$n_{i}=2$ .

$[tex]n_{f}=3,4,5,.....\infty$ .

Now the value of wavelength can be calculated as,

$\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}$ .

Here, $R=109677 cm^{-1}$ .

And $n_{f}=3$ .

Now,

$\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3})$ .

Therefore,

$\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm$

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

8 0

3 years ago

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An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire

dem82 [27]

Answer:

1,378.9ms²

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

Acceleration

Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

Hence the acceleration is 1,378.9ms²

7 0

3 years ago

Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti

Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

a). 100.2 MHz (typical frequency for FM radio broadcasting)

Then, $\lambda$ can be isolated from equation 1:

$\lambda = \frac{c}{\nu}$ (2)

since the value of c is $3x10^{8}m/s$ . It is necessary to express the frequency in units of hertz.

$\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $100200000Hz$

But $1Hz = s^{-1}$

$\nu = 100200000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}$

$\lambda = 2.99 m$

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

$\nu = 1070kHz . \frac{1000Hz}{1kHz}$ ⇒ $1070000Hz$

But $1Hz = s^{-1}$

$\nu = 1070000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}$

$\lambda = 280.3 m$

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

$\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $835600000Hz$

But $1Hz = s^{-1}$

$\nu = 835600000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}$

$\lambda = 0.35 m$

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0

3 years ago

A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy.

statuscvo [17]

Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.

According to the first law of thermodynamics,
ΔE = Q + W
178 = 658 + W
∴ W = 178-658 = -480 J

Minus sign indicates that work is done by the system.

7 0

3 years ago

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Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding. Disregarding friction, what is the velo

lilavasa [31]

Answer:

The velocity of the Mr. miles is 17.14 m/s.

Explanation:

It is given that,

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding, h = 15 m

We need to find the velocity of the Mr. Miles at the bottom of the slide. It is a case of conservation of energy which states that the total energy of the system remains conserved. Let v is the velocity of the Mr. miles. So,

$v=\sqrt{2gh}$

g is the acceleration due to gravity

$v=\sqrt{2\times 9.8\times 15}$

v = 17.14 m/s

So, the velocity of the Mr. miles is 17.14 m/s. Hence, this is the required solution.

5 0

3 years ago