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swat32
3 years ago
7

Write the sequence of filling electrons in the subshell in order of increasing energy.​

Physics
1 answer:
tankabanditka [31]3 years ago
5 0

Answer:

In a way of energy “1s” subsell contains most energy. because 1 s subsell contains the electrons of K orbital which is nearest from the nucleus. so the order of energy will be 1s>2s>2p> 3s> 3p> 4s> 3d> 4p> 5s> 4d> 5p> 6s> 4f> 5d> 6p> 7s and the increasing order of thair energies will be- 4s <3d<4p<5s<4d<5p<6s<4f.

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Algebraicallycalculate the rise time and fall timevalues for a sine wave at an arbitrary frequency f0(expressed in Hz). Assume t
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Answer:

185.95 μsec

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Given that:

Sine wave = Amsinw_{o}t

V_{min}=-V_{max}

Sinusodial signal f_{o}  = 50 kHz = 5*10^4Ht

Rise time is said to be defined as the time needed for a pulse to rise from 10% - 90% of maximum rate.

∴

t_1 = 10% = 0.1 Am

t_2 =90% = 0.9 Am

Using sine wave for  t_1; we have:

Amsinw_{o}t_1  = 0.1 Am

sinw_{o}t_1 = 0.1

w_{o}t_1 = sin^{-1}(0.1)

w_{o}t_1 = 5.7392

t_1=\frac{5.7392}{2 \pi f_0}

t_1=\frac{0.9134}{ f_0}

Using sine wave for t_2 ; we have:

Amsinw_{o}t_2  = 0.9 Am

sinw_{o}t_2  = 0.9

w_ot_2 = sin^{-1}(0.9)

w_ot_2 = 64.158

t_2 = \frac{64.1581}{2 \pi f_o}

t_2 = \frac{10.211}{f_o}

Change in rise time t_r = t_2 -  t_1

t_r = \frac{10.211}{f_o}-\frac{0.9134}{ f_0}

t_r = \frac{10.211-0.9134}{f_o}

t_r = \frac{9.2976}{f_o}

since;   f_{o}  = 50 kHz = 5*10^4Ht

t_r = \frac{9.2976}{5*10^4}

t_r = 1.85952 × 10⁻⁴

t_r = 185.952 × 10⁻⁶ sec

t_r =  185.95 μ sec

∴ The rise in time in (μ sec) for the sinosodial signal at 50 kHz = 185.95 μ sec

8 0
3 years ago
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