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2 years ago

What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state?

Physics

1 answer:

Sindrei [870]2 years ago

7 0

Answer:

$9.12\cdot 10^{-8} m$

Explanation:

The energy needed to ionize a hydrogen atom in the ground state is:

$E=13.6 eV= 2.18\cdot 10^{-18}J$

The energy of the photon is related to the wavelength by

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

Solving the formula for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2.18\cdot 10^{-18}J}=9.12\cdot 10^{-8} m$

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Assuming that each of the following objects is a typical example of its class, rank them by increasing density.

inysia [295]

molecular cloud <interstellar cloud <1 Msun protostar <1 Msun star <intercloud gas

Explanation:

Molecular cloud- They are a variety of interstellar cloud in which molecular hydrogen can sustain themselves. They have a very low temperature ranging from -440 to -370 degrees Fahrenheit or between 10 to 50 Kelvin. Owing to their extremely low temperature, they appear mostly dark when viewed through telescopes.

Interstellar cloud- They are a congregation of a large number of interstellar gases, dust and plasma in any galaxy or universe. They have varying temperature depending on their proximity to a star. E.g. Neutral hydrogen atom clouds have a temperature of around just 100 Kelvin while those in the near vicinity of a star have temperatures as high as 10,000 Kelvin.

1 Msun star- These stars have temperature anywhere between 5300 and 6000 Kelvin. The main source of such high surface temperature is nuclear fusion process where elemental hydrogen molecules are fused to form helium molecules.

1 Msun protostar- protostar is rather a young star which is still in formation phase (i.e. gathering mass from the parent molecular cloud). They have temperature anywhere between 2000-3000 kelvin and are accompanied by dust usually.

Intercloud gas- These are the remainder gases that are spread throughout the interstellar space. This Intercloud gas is divided into warm intercloud medium and extremely hot coronal gas with temperatures comparing to Sun’s corona. Warm intercloud forms the dominant part of intercloud gas with a temperature around 8000 Kelvin.

8 0

2 years ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

aleksandr82 [10.1K]

Answer:

(a) $W=-19.25J$

(b) $W=-52.8J$

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

$W=nRTln(\frac{V_1}{V_2} )$

Whereas the moles are computed from the given mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Thus, the work is:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Regards.

4 0

2 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude

bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

$\omega_{f}=\omega_{i}+\alpha t$

Put the value in the equation

$\omega_{f}=0.240+0.917\times0.203$

$\omega_{f}=0.426\ rev/s$

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of tangential speed

$v= r\omega$

Put the value into the formula

$v = \dfrac{0.720 }{2}\times0.426\times2\pi$

$v=0.963\ m/s$

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.36\times0.917\times2\pi$

$a_{c}=2.074\ m/s^2$

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0

2 years ago

A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow

leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

$F=ma$

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

$a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}$

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

7 0

2 years ago

Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri

Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

Q = 24 pC = 24 10⁻¹² C

E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

E₂ = 7.2 N / C

let's find the difference between these two fields

ΔE = E₂ -E₁

ΔE = 7.2 - 5.4

ΔE = 1.8 N / C

the minimum detection field is

E_minimum = 0.77 N / C

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0

2 years ago