What is the wavelength (in nm) of the line in the spectrum of the hydrogen atom that arises from the transition of the electron from the orbital with n =5 to the orbital with n = 2?

Note, the line corresponds to the energy of the photon that is emitted in the transition.

4.34

Find the de Broglie wavelength (in nm) associated with an electron that is moving with a velocity of 5330 km/s. The electron rest mass is 9.11 x 10-31 kg.

Note, electrons having this speed would need to be treated as waves in atoms because the wavelength is on the order of the size of atoms.

1.364e-10

Please help! I got these wrong for some reason.

4.34 is right EXCEPT you left off the 10^-7 AND you didn't convert to nm as the problem asks.

Do it again and see if you don't get 4.34E-7 m which = 4.34E-7 x (1E9 nm/1 m)=?

For #2, 1.35E-10 is correct for meters. The problem asks for nm. You need to convert to nm.

1.3646E-10 which I would round to 1.36E-10.

Also, note that 1.364 is too many significant figures; you're allowed only 3 places.

To find the wavelength of the line in the spectrum of the hydrogen atom that arises from the transition of the electron from the orbital with n = 5 to the orbital with n = 2, we can use the Rydberg formula:

1/λ = R * (1/n1^2 - 1/n2^2)

where λ is the wavelength, R is the Rydberg constant (approximated to be 1.097 x 10^7 m^-1), and n1 and n2 are the initial and final orbitals, respectively.

Plugging in the values:
n1 = 5
n2 = 2

1/λ = (1.097 x 10^7 m^-1) * (1/5^2 - 1/2^2)
1/λ = (1.097 x 10^7 m^-1) * (1/25 - 1/4)
1/λ = (1.097 x 10^7 m^-1) * (4/100 - 25/100)
1/λ = (1.097 x 10^7 m^-1) * (-21/100)

Now, we can solve for λ:

1/λ = -2.307 x 10^6 m^-1
λ = -1 / (2.307 x 10^6 m^-1)
λ ≈ -4.34 x 10^-7 m

Since the wavelength needs to be in nanometers (nm), we can convert it:

λ ≈ -4.34 x 10^-7 m * (1 nm / 10^-9 m)
λ ≈ -4.34 x 10^2 nm
λ ≈ 434 nm

Therefore, the wavelength of the line in the spectrum of the hydrogen atom that arises from the transition of the electron from n = 5 to n = 2 is approximately 434 nm.

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To find the de Broglie wavelength associated with an electron moving with a velocity of 5330 km/s, we can use the de Broglie equation:

λ = h / p

where λ is the wavelength, h is the Planck constant (approximated to be 6.626 x 10^-34 J*s), and p is the momentum of the particle.

To find the momentum, we can use the equation:

p = m * v

where m is the mass of the electron and v is its velocity.

Plugging in the values:
m = 9.11 x 10^-31 kg
v = 5330 km/s = 5330 x 10^3 m/s

p = (9.11 x 10^-31 kg) * (5330 x 10^3 m/s)
p = 4.863 x 10^-23 kg*m/s

Now, we can calculate the wavelength:

λ = (6.626 x 10^-34 J*s) / (4.863 x 10^-23 kg*m/s)
λ = 1.361 x 10^-11 m

Since the wavelength needs to be in nanometers (nm), we can convert it:

λ ≈ 1.361 x 10^-11 m * (1 nm / 10^-9 m)
λ ≈ 1.361 x 2 = 2.722 nm
λ ≈ 2.72 nm

Therefore, the de Broglie wavelength associated with an electron moving with a velocity of 5330 km/s is approximately 2.72 nm.

To determine the wavelength of the line in the spectrum of the hydrogen atom that arises from the transition of the electron from the orbital with n = 5 to the orbital with n = 2, you can use the Rydberg formula. The Rydberg formula is given by:

1/λ = R_H * (1/n_f^2 - 1/n_i^2)

where λ is the wavelength, R_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m^-1), n_f is the final energy level, and n_i is the initial energy level.

In this case, n_f = 2 and n_i = 5. Plugging these values into the formula, we get:

1/λ = 1.097 x 10^7 m^-1 * (1/2^2 - 1/5^2)
= 1.097 x 10^7 m^-1 * (1/4 - 1/25)
= 1.097 x 10^7 m^-1 * (0.25 - 0.04)
= 1.097 x 10^7 m^-1 * 0.21
= 2.3057 x 10^6 m^-1

Now, to convert this wavelength to nanometers (nm), we can use the conversion factor:
1 m = 1 x 10^9 nm

Therefore, the wavelength in nm is given by:

λ = 1/(2.3057 x 10^6 m^-1) * (1 x 10^9 nm/m)
= 1 x 10^9 nm / (2.3057 x 10^6 m^-1)
≈ 4340 nm

Hence, the wavelength of the line in the spectrum of the hydrogen atom that arises from this electron transition is approximately 4340 nm.

Now, let's move on to the second question.

In order to find the de Broglie wavelength associated with an electron moving with a given velocity, we can use the de Broglie wavelength equation:

λ = h / p

where λ is the wavelength, h is the Planck's constant (approximately 6.626 x 10^-34 J·s), and p is the momentum of the electron.

To find the momentum of the electron, we need to use the equation:

p = m * v

where p is the momentum, m is the mass of the electron, and v is the velocity.

In this case, m = 9.11 x 10^-31 kg and v = 5330 km/s.

First, we need to convert the velocity from km/s to m/s. Since 1 km = 1000 m and 1 s = 1000 ms, we have:

v = 5330 km/s * (1000 m/km) * (1 s / 1000 ms)
= 5330 m/s

Now, we can calculate the momentum:

p = (9.11 x 10^-31 kg) * (5330 m/s)
= 4.859 x 10^-27 kg·m/s

Finally, we can use the de Broglie wavelength equation to find the wavelength (λ):

λ = (6.626 x 10^-34 J·s) / (4.859 x 10^-27 kg·m/s)
= 1.363 x 10^-7 m

To convert this wavelength to nanometers (nm), we can use the conversion factor: 1 m = 1 x 10^9 nm

Therefore, the de Broglie wavelength associated with the given electron is:

λ = (1.363 x 10^-7 m) * (1 x 10^9 nm/m)
≈ 1.363 x 10^-10 nm

Hence, the de Broglie wavelength associated with the electron moving at a velocity of 5330 km/s is approximately 1.363 x 10^-10 nm.

I hope this helps you understand the correct calculations for these problems!