When a 6 Ω resistor is connected across the terminals of a 3 V battery, the number of coulombs passing through the resistor per second is

Given that - R = 6 ohm

V = 3 volt

n = ?

so by ohms law , V = I X R

I = V \ R = 3 \ 6 = 0.5 A

we know that I = Q / T here t is 1 sec

Q = I X T = 0.5 X 1 = 0.5 C

By the law of quantization ,

Q = n x e

n = Q / e = 0.5 / 1.6 x 10 ^-19

n = 0.3125 x 10^19

hoped so u got the answer mark me as brainellist :)

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m