91Ó°ÊÓ

Prob (in $\mathrm{δ}x$at x)=p(x)$\mathrm{δ}$x.........(1)

Here p(x) is the probability density and $\mathrm{δ}$x is the small width

From the figure EX39.13, the probability density of the electron at the position x=0.000 mm is as follows.

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.50 mm^-1

= 0.50x10³ m^-1

Substitute for p(x) 1x10^-5m for $\mathrm{δ}x$in the equation(1) to solve for

prob(in 0.010mm at 0.000mm)

prob(in 0.010mm at 0.000mm)=(0.50x10³m^-1)(1x10^-3m)

=5.0x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=0.000mm is 5.0x10^-3

From the figure EX39.13, the probability density of the electron at the position x=0.000 mm is as follows.

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.50 mm^-1

= 0.50x10³ m^-1

Substitute 0.50x10³ m^-1 for p(x) 1x10^-5m for $\mathrm{δ}x$in the equation(1) to solve for

prob(in 0.010mm at 0.000mm)

prob(in 0.010mm at 0.500mm)=(0.25x10³m^-1)(1x10^-3m)

=2.5x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=0.500mm is 2.5x10^-3

From the fig. the probability density of the electron at the position x=1.000mm is as follows

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.0 mm^-1

= 0.0x10³ m^-1

Substitute 0.0m^-1for p(x) 1x10^-3 for $\mathrm{δ}x$in the equation (1) to solve for

prob(in 0.010mm at 1.000mm)

prob(in 0.010mm at 1.000mm)=(1.000x10³m^-1)(1x10^-3m)

= 1.000x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=1.000mm

From the fig. the probability density of the electron at the position x=1.000mm is as follows

p(x)=0.50mm

From the fig . the probability density of the electron at the position x=0.000mm is follows.

p(x)= 0.25 mm^-1

= 0.25x10³ m^-1

Substitute 0.25x10³ m^-1 for p(x) 1x10^-3 for 1x10^-5in the equation (1) to solve for

prob(in 0.010mm at 2.000mm)

prob(in 0.010mm at 2.000mm)=(0.25x10³m^-1)(1x10^-3m)

=2.5x10^-3

Therefore the probability of electron will land in a 0.010 mm wide strong at a position of x=2.000mm is 2.5x10^-3