课后习题答案2-咨询培训-

1-9解：m=1kgv2=

10m

s-1v1

=

-

25m

s-1t

=

0.02s(1)

I

=m

(v2-

v1)

=1

[10-(-25)]

N

s

=-35N

s(2)11/15/20233习题11/15/20234（P55）1atm=760mmHg=101325pa1mmGh=133pa1pa=1N/m2

3-8一直立圆柱形容器，高0.2m,直径0.1m,顶部开启,底部有一面积为10-4m2的小孔,水以每秒1.4×10-4m3的快慢由水管自上面放入容器中.问容器内水面可上升的高度?若达到该高度时不再放水,求容器内的水流尽需多少时间.解:设顶部及底部的面积与速度分别为S1V1S2V211/15/2023511/15/2023611/15/202373-811/15/2023811/15/202395-411/15/202310《医学物理学》——第七版

胡新珉主编P253

例15-2解：

e=pEe=P2/2m

Ep=hc/11/15/202311Ver7(Chapter

15)《医学物理学》——第六版

胡新珉主编P26715-7解:λp=350

nm=3.5x10-7

m•λc=

290

nm=2.9x10-7m11/15/202312Ver7(Chapter

15)15-8解：已知

T=5800k

d=13.9x108mt=86400x365

Sec由质能方程

E=mc2

得11/15/202313Ver7(Chapter

15)15-10解：已知

n=5=500nm=5×10-7

m

t

=

1Sec11/15/202314Ver7(Chapter

15)15-11解：已知

λ0=

400

nm=4.0×10-7

m

θ=π•λ’0=

0.05

nm=5.0×10-11

m11/15/202315Ver7(Chapter

15)15-12解：已知U=9×103v

Δx=d=1mm=1×10-4

m由不确定关系:11/15/202316Ver7(Chapter

15)15-13解：已知B=0.025T

R=8.3×10-3

mq=3.2×10-19

C

mα

=6.64X10-27

Kg11/15/202317Ver7(Chapter

15)11/15/202318Ver7(Chapter

15)15-14解：已知=430nm=4.3x10-7

m11/15/202319Ver7(Chapter

15)15-16解：已知

L=10-2

m

L’=10-2

m11/15/202320Ver7(Chapter

15)15-17解：11/15/202321Ver7(Chapter

15)11/15/202322Ver7(Chapter

15)15-18解：11/15/202323Ver7(Chapter

15)15-19解：a=10-5m

m=10-14g=10-17kg

n1=100

n2=10111/15/202324Chapter

16

problems

P28311/15/202325

16-5解：U=250kv=2.5x105

vI=40mA=4x10-2

A

k=1-

0.7%=0.993t=60secQ=kIUt•

=0.993x

4x10-2

x

2.5x105

x60

(J)•

=5.958x105

J•

=595.8KJChapter

16

problems16-6解：U=80kvEmax=eU=80kev11/15/202326Chapter

16

problems

16-

X射线被衰减是,要经过几个半价层,强度才减少到原来的1%?解：

I/I0=1%=0.0111/15/202327Chapter

16

problemsm

=0.13-8解：

=3g/cm3

;

m=0.03cm2/g

;

=x1=0.1cm-1x2=0.5cm

-1x3=1cm

-111/15/202328al

=132cm-111/15/202329pb=2610cm-113-9解：

=1.54

ÅXpb=1mm=0.1cmChapter

16

problems16-10解:x=2×10-3

m

I/I0=0.2•11/15/202330addition

problem1

additional

problem

1:

The

photoelectric

threshold

wavelengcertain

metal

surface

is

2840

Å,

calculate

the

wavelength

ofrequired

to

produce

photoelectrons

whose

maximum

velocity6x105m/s.Solution:

λ0=2840

Å=2.84x10-7

m

Vmax

=

6x105m/s11/15/202331addition

problem2Additional

problem

2:

Calculate

the

wavelength

of

the

first

four

lines

of

the

Lymaatomic

hydrogen,

To

what

region

of

the

spectrum

do

these

linbelong?Solution:

Lyman

series

spectrum

is

related

to

n

1

transitwe

have

the

equation

:11/15/202332addition

problem3

Additional

problem

3:

Calculate

the

ground

state

energy

proton

(mass

1.67x10-27kg)

that

is

confined

within

a

one-

dimensional

region

1.8x10-15

m

wide)solution:

m=

1.67x10-27kg

L=

1.8x10-15

mso11/15/202333addition

problem4

Calculate

the

photon

energy

of

light

from

a

helium-Newhose

wavelength

is

632nm.Solution:

λ0=6320

Å=6.32x10-7

m

Additional

problem

4:

Calculate

the

uncertainty

in

vean