Question

velocity of a neutron with 100pm wavelength

Answer 1

@Vocaloid

Answer 2

make sure to convert pm to m first

Answer 3

but what would the mass be?

Answer 4

mass of a neutron in kg

Answer 5

okay so i woud first rearrange the equation to v=h/(m)(lambda)

Answer 6

yes

Answer 7

is the rearragement good

Answer 8

yes

Answer 9

3.956x10^-21

Answer 10

as your final solution? hm, no, a neutron would move much faster than that

100pm * (10^(-12) will convert pm to meters

so

100 * 10^(-12) = (6.626e-34)/(v * 1.674927351e-27)

solve for v

Answer 11

3656

Answer 12

hm, I got something slightly different
chucking it into a calculator gives me 3955.99 m/s (they only gave you 1 sig fig so round up to 4000?)

Answer 13

Yeah I would believe so

Answer 14

lightbulb can produce 9.8x10^20 photons in one second with a wavelength of 434nm. Hoe much energy in joules will this light build produce in one hour

Answer 15

okay so first is to convert nm to m and I got 10x10^-18

Answer 16

and the formula would be E=hv=hc/lambda

Answer 17

hm. not 100% sure but here are my thoughts:

E = hc/lambda

plugging in planck's constant, wavelength, speed of light to get E for one photon
then multiply by the number of photons 9.8x10^20 photons to get joules
then, since that's energy per second, multiply by 3600 to get energy per hour since 1 hour = 3600 seconds

Answer 18

wave length would be 10x10^-18

Answer 19

hm not quite

1 nm = 10*(-9) m

so 434 nm = 434 * 10^(-9) m

Answer 20

1.615x10^-8 energy/second

Answer 21

hm, not quite, you may have inverted something

anyway

E = hc/lambda = 6.626e-34*3e8/(434*10^(-9))

multiplying by the number of photons, then multiplying by 3600

6.626e-34*3e8/(434*10^(-9)) * 3600 * 9.8x10^20

Answer 22

for some reason I'm getting 0 x 10^20

Answer 23

hm. might be an exponent issue w/ whatever program you're using

anyway I end up with 1.6e6

Answer 24

Joules

Answer 25

yes

Answer 26

spectral lines in the Balmer series, emission back down to the n=2 level, for the hydrogen atom occur at 656.3nm, 486.1nm, 434.0nm, and 410.2nm. What is the wavelength for the next line in this series

Answer 27

hm. i'd have to think about this one for a bit. you'd probably use the rydberg equation where nf is 2 at least.

Answer 28

this would be the En=rhc/n^2?

Answer 29

hm, close but not quite. we are dealing with wavelengths not energies

so
1/lambda = R(1/nf^2 - 1/ni^2)

anyway I got what they were going for. if you let nf = 2, and plug in ni = 3,4,5, and 6, you end up with the wavelengths they give you

so the next one must be ni = 7

1/lambda = (1.097e-7)(1/2^2 - 1/7^2)

solve for lambda

Answer 30

3.x10^-9

Answer 31

3.9x10^-9

Answer 32

hm, I got something a bit different, (397nm)
i'm using wolframalpha for my calculations

Answer 33

do i vhange it to m

Answer 34

there's an unspoken rule in math that you should give back whatever units they give you

so since they give you nm in the original equation stick with nm