Explanation:
It is given that,
Focal length of the concave mirror, f = -13.5 cm
Image distance, v = -37.5 cm (in front of mirror)
Let u is the object distance. It can be calculated using the mirror's formula as :
u = -21.09 cm
The magnification of the mirror is given by :
m = -1.77
So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.
Answer:
(a) The energy of the photon is 1.632 x J.
(b) The wavelength of the photon is 1.2 x m.
(c) The frequency of the photon is 2.47 x Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
-
= -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x )
-
= 1.632 x
Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x Js), c is the speed of light (3 x
m/s) and λ is the wavelength.
10.20(1.6 x ) = (6.6 x
* 3 x
)/ λ
λ =
= 1.213 x
Wavelength of the photon is 1.2 x m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x = 6.6 x
x f
f =
= 2.47 x Hz
Frequency of the emitted photon is 2.47 x Hz.
Answer:
Rate = vmax k3/k2+k3
Explanation:
The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.
This is usually expressed as the Km ie. Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.
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