Question

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

λ (nm)
500
450
400
v × 10–5 (cm s–1)
2.55
4.35
5.35

Answer 1

plz hlp

Answer 2

total energy given by photon = Work function(minimum energy required to liberate electron) + Kinetic Energy
\[\LARGE E= \phi +K\]

Answer 3

\[\Large E = {h \cdot c \over \lambda}\]
\[\Large \phi = {h \cdot c \over \lambda_\max } \]
\[\Large K= \frac 12 mv^2\]

Answer 4

λmax = threshold wavelenghth
m = electron mass

Answer 5

bt d answer will nt come

Answer 6

wavelength of all value come different.

Answer 7

and planck constant also nt come

Answer 8

\[\large {hc \over \lambda}=\phi + \frac 12 m_e v^2\]
1) ?=500nm , v=2.55*10^5 m/s
\[\large {h \cdot (3\times10^8m/s) \over 5\times 10^{-7}m}=\phi + \frac 12 (9.11\times10^{-31}kg) (2.55\times10^{5}m/s)^2\]
\[\implies \large (6.00 \times10^{-14})h=\phi+(2.96\times10^{-20}J)\]

***Note the velocities should be in \(\Large 10^5 m \cdot s^{-1}\) for your answers to be even close to accurate

and use the 1st & 3rd pairs of data for more accurate results
.... at 450nm, the the max electron velocity should be closer to \(4.05\times10^5 m/s\)

Answer 9

bt d value of h=6.62*10^
-34.i m nt getting this one

Answer 10

using simultaneous equations:
from 3) λ=400nm v = 5.35*10^5m/s : \[\large (7.49 \times10^{14})h=\phi+(13.04\times10^{-20}J)\]
we already have: \[\large (6.00 \times10^{14})h=\phi+(2.96\times10^{-20}J)\]

subtract & we get: \[\Large (1.49 \times10^{14}s^{-1})h=(10.08\times10^{-20}J)\]

h comes out to\[\huge 6.7\times10^{-34}J \cdot s\]

Answer 11

actual λmax of sodium is around 543nm.