The answer for the problem is explained below.
The option for the answer is "D".
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
Explanation:
Given:
wavelength (λ) = 468 nm = 468×10^-9 m
speed of light (c) = 3.00 x 10^8m/s
Planck's constant is 6.626 x 10^-34J·s
To solve:
energy of light (E)
We know,
E =(h×c) ÷ λ
E = ( 6.626 x 10^-34 × 3.00 x 10^8) ÷ 468×10^-9
E = 4.25 × 10^-19 J
<u><em>Therefore the energy of the light is 4.25 × 10^-19 J</em></u>
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
C & B are switched so I'm not sure if that was a typo or not, but the answer is concentration!
Answer:
6.533 × 10^-21J
Explanation:
The energy of the microwave photon can be calculated using:
E = hf
Where;
E = energy of photon (J)
h = Planck's constant (6.626 × 10^-34 J/s)
f = frequency (9.86 x 10^12 Hz)
Hence, E = hf
E = 6.626 × 10^-34 × 9.86 x 10^12
E = 65.33 × 10^(-34 + 12)
E = 65.33 × 10^(-22)
E = 6.533 × 10^-21J
The energy of the microwave photon is
6.533 × 10^-21J
17 protons 17 electrons 18 neutrons
