The answer to this question is A
Answer:
λ = 0.002 nm
The given photon is either x-ray or gamma ray because the range of x-ray and gamma ray is 1 nm-0.1 pm.
Explanation:
Given data:
Energy of photon = 1.10 × 10⁻¹³ J.
Wavelength of photon = ?
Solution:
Formula:
E = h.c / λ
λ = h. c / E
λ = 6.626 × 10⁻³⁴ j. s × 3×10⁸ m/s / 1.10 × 10⁻¹³ J.
λ = 19.878 × 10⁻²⁶m / 1.10 × 10⁻¹³ J
λ = 18.071 × 10⁻¹³ m
λ = 18.071 × 10⁻¹³ × 10⁹
λ = 18.071 × 10⁻⁴ nm
λ = 0.002 nm
The given photon is either x-ray or gamma ray because the range of x-ray and gamma ray is 1 nm-0.1 pm.
Answer:
PART A: 412.98 nm
PART B: 524.92 nm
Explanation:
The equation below can be used for a diffraction grating of nth order image:
n*λ = d*sinθ
Therefore, for first order images, n = 1 and:
λ = d*sinθ
.
The angle θ can be calculated as follow:
tan θ = 9.95 cm/15.0 cm = 0.663 and
θ = 
(0.663) = 33.56°
Thus: d =λ/sin θ = 461/sin 33.56° = 833.97 nm
PART A:
For a position of 8.55 cm:
tan θ = 8.55 cm/15.0 cm = 0.57 and
θ = (0.57) = 29.68°
Therefore:
λ =d*sin θ = 833.97*sin 29.68° = 412.98 nm
PART B:
For a position of 12.15 cm:
tan θ = 12.15 cm/15.0 cm = 0.81 and
θ = (0.81) = 39.01°
Therefore:
λ =d*sin θ = 833.97*sin 39.01° = 524.92 nm
Answer:
consumers
Explanation:
and I you need a example it's heterotroph
