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Viefleur [7K]
3 years ago
15

Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a princi

pal quantum number of 1.
Physics
1 answer:
erica [24]3 years ago
3 0

Answer:

L_max= 0

Explanation:

The formula for magnitude of maximum orbital angular momentum is given by

L_{max}= \sqrt{l(l+1)\bar{h}}

l= orbital quantum number

l= n-1

n= shell number or principal quantum number

for n=1 , l=0

therefore, L_{max}= \sqrt{0(0+1)\bar{h}}

L_max= 0

You might be interested in
a toy dart gun generates a dart with .140kg.m/s momentum and a velocity of 4m/s. what is the mass of the dart in grams?
sleet_krkn [62]

Answer: m = 0.035kg = 35g

Explanation: Momentum p=0.140kgm/s

Velocity v=4m/s

Mass m=?

Formula-

Momentum depends on the mass of the object in motion and its velocity.

The equation for momentum is

p = mv

m = p/v

m = 0.140/4

m = 0.035kg

m = 35g

Hence, in the toy dart gun mass of the dart is 0.035kg.

4 0
2 years ago
At sunset, red light travels horizontally through the doorway in the western wall of your beach cabin, and you observe the light
Nady [450]

Answer:

9.8\cdot 10^{-6}m

Explanation:

For light passing through a single slit, the position of the nth-minimum from the central bright fringe in the diffraction pattern is given by

y=\frac{n \lambda D}{d}

where

\lambda is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have

\lambda=700 nm = 7.00\cdot 10^{-7}m is the wavelength of the red light

D = 14 m is the distance of the screen from the doorway

d = 1.0 m is the width of the doorway

Substituting n=1 into the equation, we find the distance between the central bright fringe and the first-order dark fringe (the first minimum):

y=\frac{(1)(7.00\cdot 10^{-7} m)(14 m)}{1.0 m}=9.8\cdot 10^{-6}m

6 0
3 years ago
Please answer any of these thanks !
KIM [24]
1).  The equation is: (speed) = (frequency) x (wavelength)

Speed = (256 Hz) x (1.3 m) = 332.8 meters per second

 2).  If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.

If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.

If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.

3).  The equation is:  Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)

Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.

That's ' 200 k Hz ' .

Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised.  They broadcast on several different frequencies,
and one of them is 198 kHz !
7 0
3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

4 0
3 years ago
Electromagnet Fluctation
Tems11 [23]

Answer:

answer choice B

Explanation:

6 0
3 years ago
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